Class 8 Maths - Exponents And Powers

Evaluate : $\left (\dfrac 8{125} \right )^{-\tfrac13}$

Here we have,

$\ \left (\dfrac8{125} \right )^{-1/3}$

$= \left [ \dfrac{\left (2 \times 2 \times 2 \right )}{ \left (5 \times 5 \times 5 \right )} \right ]^{-1/3}$

$= \left (\dfrac{2^{3}}{5^{3}} \right )^{-1/3}$

$= \left (\dfrac 25 \right )^{3 \times -1/3}$

$= \left (\dfrac 25 \right )^{-1}$

$= \dfrac 52 = 2\space \dfrac12$

$\left ( \dfrac{-1}{27} \right )^{-2/3}$

Here we have,

$\left ( \dfrac {-1}{27} \right )^{-2/3}\\= \left (\dfrac{-1^3}{3^{3}} \right)^{-2/3}\\ = \left (\dfrac{-1}3 \right )^{3 \times -\tfrac23}\\ = \left (-\dfrac 13 \right )^{-2}\\ = \left (-3 \right )^{2}= 9$

Identify the greater number, wherever possible, in the following.
$2^{10}$ or $10^{2}$ .

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Express the number appearing in the following statement in standard form.
The earth has $1,535,000,000$ cubic $km$ of seawater.

2123007_1218720_ans_fe29b06c11134310a6251f197ae8dd8d.jpg

Express the number appearing in the following statement in standard form.
Diameter of the Earth is $1,27,56,000$ meters.

2123515_1218696_ans_ddd2b0b7e55e4974b91f442c260c6e31.jpg

Express the the mean distance between Earth and moon is $384,000,000\ m$ in standard from.

Standard form of

$3,84,000,000$ is

$3.84 \times 10^{8}$

Express the number appearing in the following statement in standard from.
The univers is estimated to be $12,000,000,000$ years old.

1350213_1218703_ans_a44627f6ee9e4af5acbbfed813f261fa.jpeg

Express the number appearing in the following statement in standard from:
The distance between Sun and Saturn is $1,433,000,000,000\ m\ .$

Standard form of

$1,433,000,000$ is

$1.433 \times 10^{9}$ .

Write the given product in exponential form.
$7\times a\times a\times a..... 8$ times $\times b\times b\times b\times … 5$ times.

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Write the given expression in the product form.
$30x^4y^4z^5$

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Write the given product in exponential form.
$4\times a\times a\times ….. 5\text{ times }\times b\times b\times ….. 12\text{ times }\times c\times c..... 15\text{ times }$ .

2123533_1674303_ans_d2635ba01f2349309412d4e5fa1323d7.jpg

Express the numbers appearing in the following statements in the standard form:
The universe is estimated to be about 12,000,000,000 years old.

2122955_1676692_ans_4c1140f67e1d4fb1bff948998632a0cb.jpeg

Express the numbers appearing in the following statements in the standard form:
Diameter of the Earth is 1,27,56,000 metres.

2124528_1676679_ans_da939a44b95e4ed5b5478b91e785f928.jpeg

Evaluate $64^{\tfrac{-1}{2}}$ .

1680411_1715586_ans_1a74c1f3d6494ecbb0c5c800c8a44936.jpg

Identify the greater number, wherever possible, in the following?
$100^2$ or $2^{100}$ .

${ 100^{ 2 } }\, \, or\, \, { 2^{ 100 } } \\ \Rightarrow { 100^{ 2 } }=100\times 100=10,000 \\ \Rightarrow { 2^{ 100 } }=2\, \, hundred\, \, times \\ \Rightarrow kb{ 2^{ 10 } }=1024 \\ \Rightarrow { 2^{ 20 } }=1024\times 1024>10,000 \\ \therefore { 2^{ 100 } }\, \, is\, \, greater\, \, than\, \, { 100^{ 2 } }\, \, \, \, Ans.$

Identify the greater number in the following.
$2^8$ or $8^2$ .

2123516_1144474_ans_e1c342d88e694a79b4f9e17172c91f65.jpg

Write the following number in standard form:
$\dfrac{1}{1000000}$ .

2124203_1395270_ans_b0daf7548e1f4a7b965013d63f9abdca.jpg

Evaluate
$\left(\dfrac 3 5\right)^{-2}$

Let us evaluate the given expression,

By using the formula

$\left(a^{-n}=1/a^n\right)$

So,

$\left(\dfrac 35\right)^{-2}=\left(\dfrac 53\right)^2$

$=\left(\dfrac 53\right)\times \left(\dfrac 53\right)$

$=\dfrac{25}9$

Express the number appearing in the following statement in standard from.
$60,230,000,000,000,000,000,000$ molecules are present in a drop of water weighing $1.8\ gm$ .

1190355_1218710_ans_6354ef79abb74059935d975cd6a6ffdb.jpeg

Which is greater?
$\left (\dfrac{1}{2} \right )^{\frac{1}{2}}$ or $\left (\dfrac{2}{3} \right )^{\frac{1} {3}}$

L.C.M of

$2$ and

$3$ is

$6$ .

$\therefore \left (\cfrac{1}{2} \right )^{\tfrac{1}{2}}=\sqrt[2]{\cfrac{1}{2}}=\sqrt[2\times 3]{\left (\cfrac{1}{2} \right )^{3}}=\sqrt[6]{\cfrac{1}{8}}$

and

$\left (\cfrac{2}{3} \right )^{\tfrac{1}{3}}=\sqrt[3]{\cfrac{2}{3}}=\sqrt[3\times 2]{\left (\cfrac{2}{3} \right )^{2}}=\sqrt[6]{\cfrac{4}{9}}$

But

$\cfrac{4}{9}> \cfrac{1}{8}$

$\therefore \sqrt[3]{\cfrac{2}{3}} >\sqrt[2]{\cfrac{1}{2}}$

Prove that:
$(\sqrt{3 \times 5^{-3}} \div \sqrt [ 3 ]{ 3^{ -1} } \sqrt{5}) \times \sqrt [ 6 ]{ 3 \times 5^6 } = \dfrac{3}{5}$

Given,

$\dfrac{\sqrt{3\times 5^{-3}}}{\sqrt[3]{3}^{-1}\sqrt{5}} \times \sqrt[6]{3\times 5^6}$

$=\dfrac {3^{\dfrac{1}{2}}\times 5^{\dfrac{-3}{2}}} {3^{\dfrac{-1}{3}} \times 5^{\dfrac{1}{2}}} \times 3^{\dfrac{1}{6}}\times 5^{\dfrac{6}{6}}$

$=3^{\dfrac{5}{6}}\times 5^{-2} \times 3^{\dfrac{1}{6}}\times 5^{1}$

$=3\times 5^{-1}$

$=\dfrac{3}{5}$

$\dfrac{\sqrt{3\times 5^{-3}}}{\sqrt[3]{3}^{-1}\sqrt{5}} \times \sqrt[6]{3\times 5^6}=\dfrac{3}{5}$

Express $16^{-2}$ as a power with base $4$

2159639_616709_ans_4d8fdf1b49b243d2804be657a8b774a5.jpg

Which is greater $7^{92}$ or $8^{91}$ ?

$\dfrac{(7^{92})}{(8^{91})}$

$=(\dfrac{7}{8})^{91}\times 7$

Now, $\dfrac{ 7}{8} =0.875$

As we know that, $0.875\times0.875 \approx 0.75 (0.9 \times 0 .9 =0.81)$

$0.875\times0.75 \approx 0.6$

Next, $0.875 \times 0.6 \approx 0.5$

The point is every subsequent multiplication of $0.875$ reduces the numerator by digit

clearly after 4 more such multiplication, we land at 0.1

Therefore, the total numerator $=0.1\times7=0.7$

Which is already less than 1

Hence $(8)^{91} > 7^{92}$

Which is larger: ${ \left( 100 \right) }^{ 4 }$ or ${ \left( 125 \right) }^{ 3 }$ ?

2122809_557988_ans_aed082783d43409aaf6eb5440ab31bbb.jpg

Simplify and give reasons
$\left(\dfrac{3}{4}\right)^{-3}$

$\left(\dfrac 34\right)^{-3}\\ a^{-n}=\dfrac 1{a^n}\\\implies\left( \dfrac43 \right) ^3\\\left(\dfrac ab\right)^n=\dfrac {a^n}{b^n}\\\implies \dfrac {4^3}{3^3}=\dfrac{64}{27}$

Express the following in standard from:
i) $76854000\times 10^{8}$
ii) $0.000089\times 10^{6}$

$(i)$

$76854000 \times {10^8} \Rightarrow 7.6854 \times {10^{15}}$

$(ii)$

$0.000089 \times {10^6} \Rightarrow 8.9 \times {10^1}$

Simplify $\displaystyle \frac { \left( 5 ^ { - 3 } \right) ^ { 2 } \times 3 ^ { 4 } } { \left( 3 ^ { - 2 } \right) ^ { - 3 } \times \left( 5 ^ { 3 } \right) ^ { - 2 } }$

Let

$a\displaystyle=\frac{(5^{-3})^2\times 3^4}{(3^{-2})^{-3}\times (5^3)^{-2}}$

$a\displaystyle=\frac{(5^{-3})^2\times 3^4}{(3^{-2})^{-3}\times (5^3)^{-2}}=\frac{5^{-6}\times 3^4}{3^6\times 5^{-6}}$

$a\displaystyle=\dfrac{3^4}{3^6}$

$a=\dfrac {1}{3^2}$

$a=\dfrac{1}{9}$

$\displaystyle \therefore \frac{(5^{-3})^2\times 3^4}{(3^{-2})^{-3}\times (5^3)^{-2}}=\dfrac{1}{9}$

Compare the following numbers:
$2.7\times{10}^{12};1.5\times{10}^{8}$

$\textbf{Step 1: Given}$

Two numbers in standard form are

$2.7\times 10^{12}$ and

$1.5\times 10^8.$

$\textbf{Step 2: To Find}$

To compare the given numbers.

$\textbf{Step 3: Calculation}$

Let us make the powers of

$10$ equal in both the numbers to compare

$2.7\times 10^{12}=2.7\times 10^4\times 10^8$

$=27000\times 10^8$

Now, as the exponent of 10 in the two numbers is same we hve,

$\begin{array}{l}27000 > 1.5\\ \Rightarrow 27000 \times {10^8} > 1.5 \times {10^8}\\ \Rightarrow 2.7 \times {10^{12}} > 1.5 \times {10^8}\end{array}$

$\textbf{Step 4: Conclusion}$

$2.7 \times 10^{12} >1.5 \times 10^{8}$

Simplify and write the answer in exponential form
${ \left( -4 \right) }^{ -3 }\times { \left( 5 \right) }^{ 3 }\times { \left( -5 \right) }^{ -3 }$

$(-4)^{=3} \times (5)^3 \times (-5)^{-3}= \dfrac{-1}{4^3} \times 5^3 \times \left(\dfrac{-1}{5^3}\right) = \dfrac{1}{4^3} = 4$
1226991_1398609_ans_02a29b2290324b8a803c276acfce6692.jpg

Evaluate $\left\{\left(\dfrac{4}{3}\right)^{-1}-\left(\dfrac{1}{4}\right)^{-1}\right\}^{-1}$

$\left\{\left(\dfrac{4}{3}\right)^{-1}-\left(\dfrac{1}{4}\right)^{-1}\right\}^{-1}$

$=\left\{\dfrac{3}{4}-4\right\}^{-1}=\left\{\dfrac{3-16}{4}\right\}^{-1}=\left(\dfrac{-13}{4}\right)^{-1}=\dfrac{-4}{13}$ .

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Express $512$ in power notation.

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Prove that:
${ \left\{ { \left( \dfrac { 1 }{ 3 } \right) }^{ -1 }-{ \left( \dfrac { 1 }{ 4 } \right) }^{ -1 } \right\} }^{ -1 } = -1$

$\left\{\left(\dfrac{1}{3}\right)^{-1} -\left(\dfrac{1}{4}\right)^{-1}\right\}$

$=\left\{ 3^{1}- 4^{1}\right\}$

$=\left\{3- 4\right\}$

$=-1$ .

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Say true or false and justify your answer:
${2}^{3}> {5}^{2}$

$\textbf{Step 1: Given}$

$2^3>5^2$

$\textbf{Step 2: To do}$

To check if

$2^3>5^2$ is true or false

$\textbf{Step 3: Calcualtion}$

$\textbf{The exponent of a number says how many times to use the number in a multiplication.}$

$LHS$

$2^3=2\times 2\times 2$

$=8$

$RHS$

$5^2=5\times 5$

$=25$

We know that

$8<25$

i.e.,

$2^3<5^2$

Hence the given statement is false.

$\textbf{Step 4: Conclusion}$

Statement

$2^3>5^2$ is false.

Evaluate:
$\left[\left(-\dfrac{2}{3}\right)^{-2}\right]^3\times \left(\dfrac{1}{3}\right)^{-4}\times 3^{-1}\times \dfrac{1}{6}$

$\left[\left(-\dfrac{2}{3}\right)^{-2}\right]^3\times \left(\dfrac{1}{3}\right)^{-4}\times 3^{-1}\times \dfrac{1}{6}$

$\left[\left(-\dfrac{3}{2}\right)^{2}\right]^3\times 3^{4}\times \dfrac{1}{3}\times \dfrac{1}{3\times 2}$

$\left(-\dfrac{3}{2}\right)^6\times (3)^2\times \dfrac{1}{2}$

$=\dfrac{3^{6+2}}{2^{6+1}}$

$=\dfrac{3^8}{2^7}$

Find:
$125^{\frac{-1}{3}}$

$125^{\frac{-1}{3}} = \frac{1}{125^{\frac{1}{3}}}$

$=\dfrac{1}{(5 \times 5 \times 5)^{\frac{1}{3}}} =\dfrac{1}{(5^3)^{\frac{1}{3}}}$

$=\dfrac{1}{5^{3 \times \frac{1}{3}}} =\dfrac{1}{5}$

Evaluate
$(-3)^{-3}$

Let us evaluate the given expression,

By using the formula

$(a^{-n}=1/a^n)$

So,

$(-3)^{-3}=(-1/3)^3$

$=(-1/3)\times (-1/3)\times (-1/3)$

$=-1/27$

Evaluate:
(i) ${\left (\dfrac{1}{3}\right )^{-1} - \left (\dfrac{1}{4}\right )^{-1}}$
(ii) $\left (\dfrac{5}{8}\right )^{-7}\times \left (\dfrac{8}{5}\right )^{-4}$

(i)

${\left (\dfrac{1}{3}\right )^{-1} - \left (\dfrac{1}{4}\right )^{-1}}$

$= [3-4]^{-1} = (-1)$

(ii)

$\left (\dfrac{5}{8}\right )^{-7}\times \left (\dfrac{8}{5}\right )^{-4}$

$= \left (\dfrac{8}{5}\right )^{7}\times \left (\dfrac{5}{8}\right )^{4}$

$= \dfrac{512}{125}$

Which is greater $(150)^{300}$ or $(20000)^{100}\times (100)^{100}$ ?

$(20000)^{100}\times (100)^{100}=2^{100}\times (100)^{200}\times (100)^{200}\times (100)^{100}=(2^{1/3})^{300}\times (100)^{300}$

$\Rightarrow (20000)^{100}\times (100)^{100}=(100\sqrt[3]{2})^{300}$

$100(\sqrt[3]{2})< 150$

$\therefore \boxed{(150)^{300}}$ is greater.

Evaluate:
(i) $3^{-2}$ ii) $(-4)^{-2}$ $\left (\dfrac{1}{2}\right)^{-5}$

(i) To evaluate:

$3^{-2}$

Now,

$3^{-2} = \dfrac{1}{3^{2}} =\dfrac{1}{9}$

Hence, the required value is

$\dfrac{1}{9}$ .

(ii) To evaluate:

$(-4)^{-2}$

Now,

$(-4)^{-2} = \dfrac{1}{(-4)^{2}} =\dfrac1{4^2} = \dfrac{1}{16}$

Hence, the required value is

$\dfrac{1}{16}$ .

(iii) To evaluate:

$\left(\dfrac{1}{2}\right)^{-5}$

Now,

$\left ( \dfrac{1}{2}\right )^{-5} = \dfrac{1}{(2)^{-5}} = 2^5 = 32$

Hence, the required value is

$32$ .

Show that $(1.1)^{10000} > 1000$

Lets take log on both sides, since the values

$1.01$ &

$1000$ are greater than 1 the inequality remains same.

$\implies \log{1.01^{1000}} > \log{1000}$

$\implies 1000\times \log 1.01 > 3$

$\implies 1000\times 0.0043 > 3$

$\implies 4.3 >3$

Which is true and therefore we can say that

$(1.01)^{1000}>1000$

Find $x$ in the expression $(18)^{3.5}\div (27)^{3.5}\times 6^{3.5}=2^{x}$

$\dfrac{18^{3.5}}{27^{3.5}}\times 6^{3.5}=(\dfrac{18\times 6}{27})^{3.5}= (4)^{3.5}= 2^7=2^x$

$x=7$

Mass of the earth is $5.97\times {10}^{24}\ kg$ and mass of the moon is $7.35\times {10}^{22}\ kg$ . What is the difference of their masses?

Mass of Earth =

$= 5.97 \times {10^{24}}\,Kg$

Mass of the moon

$= 7.35 \times {10^{22}}\,Kg$

Difference of their masses in

$Kg$

$= 5.97 \times {10^{24}} - 7.35 \times {10^{22}}$

$= \left( {5.97 \times {{10}^2} - 7.35} \right){10^{22}}$

$= \left( {597 - 7.35} \right) \times {10^{22}}$

$= 589.65 \times {10^{22}}$

$= 589.65 \times {10^{24}}$

$= 5.89 \times {10^{24}}$

Compare $7 \times 10^{-6}$ and $129 \times 10^{-7}.$

First we will rewrite

$129\times10^{-7},$

$\begin{aligned}{}129 \times {10^{ - 7}} &= \frac{{129}}{{10}} \times {10^{ - 6}}\\ &= 12.9 \times {10^{ - 6}}\end{aligned}$

Now, on comparing

$12.9\times10^{-6}$ and

$7\times10^{-6}$ we get that

$10^{-6}$ is common in both and

$12.9>7.$

Hence,

$12.9\times10^{-6}>7\times10^{-6}$ or

$129\times10^{-7}>7\times10^{-6}$

Which one is greater between $31^{14}$ and $17^{18}$ ?

$We\quad know\quad that\quad 32>31\\ \rightarrow { \left( 2 \right) }^{ 5 }>31\\ Raising\quad both\quad sides\quad to\quad power\quad of\quad 14\\ \therefore \left( { \left( 2 \right) }^{ 5 } \right) ^{ 14 }>{ 31 }^{ 14 }\\ \rightarrow { 2 }^{ 70 }>{ 31 }^{ 12 }.....\left( 1 \right) \\ Now\quad consider\quad 17>16\\ \rightarrow 17>{ \left( 2 \right) }^{ 4 }\\ Raising\quad both\quad sides\quad to\quad power\quad of\quad 18\\ \therefore { 17 }^{ 18 }>\left( { \left( 2 \right) }^{ 4 } \right) ^{ 18 }\\ \rightarrow { 17 }^{ 18 }>{ 2 }^{ 72 }.....\left( 2 \right) \\ Now\quad { 2 }^{ 72 }>{ 2 }^{ 70 }....\left( 3 \right) \quad \\ From\quad points\quad 1,2\quad and\quad 3\\ { 17 }^{ 18 }>{ 31 }^{ 14 }$

Express the following in the decimal form:
$4.67\times {10}^{4}$

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Simplify: $\left(\dfrac{81}{16}\right)^{-3/4} \times \left(\dfrac{25}{9}\right)^{-3/2} \times \left(\dfrac{2}{5}\right)^{-3}$

$\left(\dfrac{81}{16}\right)^{-3/4} \times \left(\dfrac{25}{9}\right)^{-3/2} \times \left(\dfrac{2}{5}\right)^{-3}$

$=\left(\dfrac{3^4}{2^4}\right)^{-3/4} \times \left(\dfrac{5^2}{3^2}\right)^{-3/2} \times \left(\dfrac{2}{5}\right)^{-3}$

$=\left(\dfrac{3}{2}\right)^{4\times-3/4} \times \left(\dfrac{5}{3}\right)^{2\times-3/2} \times \left(\dfrac{2}{5}\right)^{-3}$

$=\left(\dfrac{3}{2}\right)^{-3} \times \left(\dfrac{5}{3}\right)^{-3} \times \left(\dfrac{2}{5}\right)^{-3}$

$=\left(\dfrac{30}{30}\right)^{-3}$

$=1$ .

Find the value of $x$ for which ${ \left\{ { \left( \dfrac { -2 }{ 7 } \right) }^{ 2 } \right\} }^{ x }\times { \left( \dfrac { -7 }{ 2 } \right) }^{ -1 }=\dfrac { -8 }{ 343 }$ .

${\left\{ {{{\left( {\dfrac{{ - 2}}{7}} \right)}^2}} \right\}^x} + {\left( {\dfrac{{ - 7}}{2}} \right)^{ - 1}} = \dfrac{{ - 8}}{{343}}$

$\Rightarrow {\left( {\dfrac{{ - 2}}{7}} \right)^{2x}} + \left( {\dfrac{{ - 2}}{7}} \right) = {\left( {\dfrac{{ - 2}}{7}} \right)^3}$

$\Rightarrow {\left( {\dfrac{{ - 2}}{7}} \right)^{2x + 1}} = {\left( {\dfrac{{ - 2}}{7}} \right)^3}$

$\Rightarrow 2x + 1 = 3$

$\Rightarrow 2x = - 2$

$\Rightarrow x = - 1$

If $9^{x} \times 3^{2}\times \left(3^{\left(\tfrac{-x}{2}\right)}\right)^{-2}=\dfrac{1}{27}$ , then find the value of $x$ .

The given equation is :

${ 9 }^{ x }\times 3^{ 2 }\times { \left( { 3 }^{ \frac { -x }{ 2 } } \right) }^{ -2 }=\frac { 1 }{ 27 }$

It can be written as:

$3^{ 2x }\times 3^{ 2 }\times 3^{ x }=3^{ -3 }$

${ 3 }^{ 2x+2+x }={ 3 }^{ -3 }$

${ 3 }^{ 3x+2 }={ 3 }^{ -3 }$

Now,since the bases are same,then comparing the powers,we get

$3x+2=-3$

$x=\frac { -5 }{ 3 }$

If $\left(\dfrac{5}{3}\right)^{-15} \times \left(\dfrac{5}{3}\right)^{11} =\left(\dfrac{5}{3}\right)^{8x}$ , then $x$ =?

2122857_1042806_ans_4923d2cbb2294fc8808fac0f39a99bed.jpeg

If the standard form of $12,000,000,000$ is $1.2\times 10^{m}$ then find the value of $m$ .

2127672_705091_ans_322afb99a2754aefa8e9b45c9d39ea16.jpg

If ${ \left( 64 \right) }^{ x }=\cfrac { 1 }{ { \left( 256 \right) }^{ y } } =2\sqrt { 2 }$ ; then show that $3x+4y=0$ .

Given

${(64)}^x=2^{1.5}$

$\Rightarrow { { (2 }^{ 6 }) }^{ x }={ 2 }^{ 1.5 }\\$

Exponent and bases should be equal

$\Rightarrow 6x=1.5$

$\Rightarrow x=0.25$

Given

${(256)}^{-y}=2^{1.5}$

$\Rightarrow { { (2 }^{ 8 }) }^{ -y }={ 2 }^{ 1.5 }\\$

Exponent and bases should be equal

$\Rightarrow -8y=1.5$

$\Rightarrow y=-\dfrac3{16}$

Now

$3x+4y=3(0.25)+4\left(-\dfrac3{16}\right) = \dfrac34 - \dfrac34$

$=0$

If $2^{x-1}+2^{x+1}=1280$ , then find the value of $x$ .

2123420_1115507_ans_e0e108a761864d62a1b0f2ec56d3e2c2.jpeg

Identify the greater number, wherever possible, in the following?
$5^3$ or $3^5$ .

${ 5^{ 3 } }\, \, or\, \, { 3^{ 5 } } \\ \Rightarrow { 5^{ 3 } }=5\times 5\times 5=125 \\ \Rightarrow { 3^{ 5 } }=3\times 3\times 3\times 3\times 3=243 \\ \therefore { 3^{ 5 } }\, \, >\, \, \, { 5^{ 3 } }\, \, \,$

If $x^{1/p} = y^{1/q} = z^{1/r}$ and $xyz = 1,$ then the value of $p+q+r = ?$

$x^{1/p}=y^{1/q}=z^{1/r}=k$ (say)

$\Rightarrow x=k^p, y=k^q, z=k^r$

$xyz=1$ (given)

$k^p.k^q.k^r=1$

$k^{(p+q+r)}=1$ .

$\boxed{\Rightarrow (p+q+r)=0}$ .

Identify the greater number between the following:
$4^3$ or $3^4$

${ 4^{ 3 } }\, \, or\, \, { 3^{ 4 } } \\ \Rightarrow { 4^{ 3 } }=4\times 4\times 4=164 \\ \Rightarrow { 3^{ 4 } }=3\times 3\times 3\times 3=81 \\ \therefore { 3^{ 4 } }\, \, >\, \, \, { 4^{ 3 } }\, \, \,$

Which is greater among the following?
$\sqrt{2}, \sqrt[3]{4}$ and $\sqrt[4]{3}$

$\sqrt 2, (4)1/3, \Rightarrow LCM\ (2, 3, 4)=12$

$(i)\ \sqrt 2=(2)^{1/2}=2^{(6/12)}=(2^6)^{1/12}=(64)^{1/12}$

$(ii)\ 4^{1/3}=(4)^{4/12}=(256)^{1/12}$

$(iii)\ 3^{1/4}=3^{3/12}=(3^3)^{1/12}= (27)^{1/12}$

$(4)^{1/3} > (2)^{1/2} > 3^{1/4}$

$\begin{array}{l} If\, \, { e^{ \left( { { { \sin }^{ 2 } }x+{ { \sin }^{ 2 } }x+{ { \sin }^{ 6 } }x......\infty } \right) \log { e^{ 2 } } } }\, and\, { y^{ 2 } }-5y+4=0\, satisfy\, the\, equation\, then\, find\, the\, \\ value\, of\, \frac { { \sin x } }{ { \cos x-\sin x } } \end{array}$

$\begin{array}{l} Given \\ { e^{ \left( { { { \sin }^{ 2 } }x+{ { \sin }^{ 2 } }x+{ { \sin }^{ 6 } }x......\infty } \right) \log { e^{ 2 } } } } \\ ={ e^{ \log 2\left( { \frac { 1 }{ { 1-{ { \sin }^{ 2 } }x } } } \right) } } \\ ={ 2^{ \left( { \frac { 1 }{ { 1-2{ { \sin }^{ 2 } }x } } } \right) } }={ 2^{ { { \sec }^{ 2 } }x } }=y \\ Now, \\ { y^{ 2 } }-5y+4=0 \\ \Rightarrow { y^{ 2 } }-4y-y+4=0 \\ \Rightarrow y\left( { y-4 } \right) -1\left( { y-4 } \right) =0 \\ \Rightarrow \left( { y-4 } \right) \left( { y-1 } \right) =0 \\ \therefore y=4\, and\, y=1 \\ If\, { 2^{ { { \sec }^{ 2 } }x } }=1 \\ { \sec ^{ 2 } }x=0\, \, \left( { not\, possible\, as\, { { secx } }\ge 1 } \right) \\ else, \\ { 2^{ { { \sec }^{ 2 } }x } }=4 \\ { \sec ^{ 2 } }x=2 \\ \cos x=\frac { 1 }{ { \sqrt { 2 } } } \\ Thus,\, the\, value\, of\, \frac { { \sin x } }{ { \cos x-\sin x } } =\frac { { \frac { 1 }{ { \sqrt { 2 } } } } }{ { \frac { 1 }{ { \sqrt { 2 } } } -\frac { 1 }{ { \sqrt { 2 } } } } } \, \Rightarrow Not\, defined. \\ \end{array}$

Write the base and the exponent of $(7x)^{2}$

Given the number is

$(7x)^2$ .

So the base is

$7x$ and the exponent is

$2$ .

Change the given number into it's standard form
$128000000$

$128000000$

$=128\times {10}^{6}$

$=12.8\times 10\times {10}^{6}$

$=1.28\times 10\times 10\times {10}^{6}$

$=1.28\times {10}^{8}$

$(\sqrt [ 5 ]{ 8 } )^{\dfrac{5}{2}}\times (16)^{\dfrac{-3}{2}}$

We have,

${{\left( \sqrt[5]{8} \right)}^{\frac{5}{2}}}\times {{\left( 16 \right)}^{\frac{-3}{2}}}$

$\Rightarrow {{\left( 8 \right)}^{\frac{1}{5}\times \frac{5}{2}}}\times {{\left( {{2}^{4}} \right)}^{\frac{-3}{2}}}$

$\Rightarrow {{\left( {{2}^{3}} \right)}^{\frac{1}{2}}}\times {{\left( 2 \right)}^{-6}}$

$\Rightarrow {{2}^{\frac{3}{2}-6}}$

$\Rightarrow {{2}^{\frac{-9}{2}}}$

Hence, this is the answer.

Express the following as a product of prime factors in exponential form :
$729\times 64$

$729 \times 64$

$=3\times 3\times 3\times 3\times 3\times 3 \times 2\times 2\times 2\times 2\times 2\times 2$

$=3^6 \times 2^6$

1771764_1595758_ans_c465ce418c2040b89989981cad8105d9.PNG

Identify the greater number in each of the following numbers:
${2}^{8}$ or ${8}^{2}$

We have : (1)

${2}^{8}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=64\times 4 = 256$

(2)

${8}^{2}=8\times 8=64$

Since

$256> 64$
Thus,

${2}^{8}$ is greater than

${8}^{2}$ .

Identify the greater number among the two numbers:
${2}^{10}$ or ${10}^{2}$

${2}^{10}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=1,024$

${10}^{2}=10\times 10=100$
Since,

$1,024> 100$
Thus,

${2}^{10}> {10}^{2}$

Identify the greater number in the following numbers:
${100}^{2}$ or ${2}^{100}$

1338811_1646428_ans_3c7233746012489e9a8b2dcd3b668158.jpeg

Express the number appearing in the following statements in standard form:
$602,300,000,000,000,000,000,000$ molecules are contained in a drop of water weighing $1.8gm$ .

2126821_1646517_ans_ea3db53a20c043c69f1af917034bdeb5.jpg

Write $629000$ in standard form.

$629000=6.29\times 100000$

$=6.29\times 10^5$

Express the number appearing in the following statement in standard form:
The Earth has $1,353,000,000$ cubic km of seawater.

2122806_1646518_ans_fb0b6e61ae2b41a783b41499d03451a4.jpg

Identify the greater number, wherever possible, in each of the following:
${5}^{3}$ and ${3}^{5}$

2122871_1646426_ans_2b650dc0f38040eaa539be9867cb9e2a.jpeg

On solving the equation $7^{2x+3} = 1$ , we get $x=\dfrac{-3}{b}$ then find the value of $b$ .

Given,

$7^{2x+3}=1$

The above equation can also be written as:

$7^{2x+3}=7^0$

By comparing powers we get;

$2x+3=0$

$\therefore x=-\dfrac{3}{2}$

Comparing the value of x with the question we got

$\Rightarrow b=2$

Standard from of $6020000000000000$ is $6.02\times 10^{16}$ .
Enter 1 if it is True or 0 if it is False

The correct answer is 0

1872099_1671200_ans_d4909c69cdc4466a866e80cb6f06376c.jpg

Express the following numbers in standard from:
$4 \div 100000$

2123374_1671229_ans_e593e15fb433482d8aaf8580f7e5b141.jpeg

Write the given expression in the product form.
$43p^{10}q^5r^{15}$

2126813_1674309_ans_e97f13578e0942e9b4d8ef927820584d.jpg

Write the given product in exponential form.
$8\times b\times b\times b\times a\times a\times a\times a$ .

We know that if a number say

$b$ is multiplied three times. That is,

$b\times b\times b$ can be written as

$b^{3}$ .

Similarly,

$8\times b\times b\times b =$

$8^{1} \times$

$b^{3}$ .

Thus

$8\times b\times b\times b\times a\times a\times a\times a =$

$8^{1} \times$

$b^{3}$

$\times a^{4}$ .

$=8^{1}b^{3}a^{4}$ .

Write the expression $5xy\times 3x^2y\times 7y^2$ in exponential form.

We know that if a number say

$b$ is multiplied three times. That is,

$b\times b\times b$ can be written as

$b^3$ .

$b^3$ =

$b\times b\times b$

Similarly,

$5xy = 5\times x \times y$ .

$3x^{2}y = 3\times x\times x\times y$ .

$7y^{2} = 7 \times y \times y$ .

Thus,

$5xy \times 3 x^{2} y \times 7y^{2} = 5 \times x \times y \times 3 \times x \times x \times y \times 7 \times y \times y$

$= 5 \times 3 \times 7 \times x \times y \times x \times x \times y \times y \times y$ .

$= 105 \times x^{3} \times y^{4}$ .

$= 105 x^{3}y^{4}$ .

Write the given expression in the product form.
$a^2b^5$ .

We know that if a number say

$b$ is multiplied three times. That is,

$b\times b\times b$ can be written as

$b^3$ .

$\implies b^3$

$=b\times b\times b$

So,

$a^{2} = a \times a$

$b^{5} = b \times b \times b \times b \times b$

Thus,

$a^{2}b^{5} = a \times a\times b\times b\times b\times b\times b$ .

Write the given expression in the product form.
$17p^{12}q^{20}$ .

Let,

$a$ number is multiplied three times.

That is,

$a^{3}$

$= a\times a \times a$ .

So,

$p^{12} = p \times p \times p \times p \times p \times p \times p \times p \times p \times p\times p\times p$ .

$q^{20} = q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q$ .

Thus,

$17p^{12}q^{20}= 17 \times p \times p \times p \times p \times p \times p \times p \times p \times p \times p\times p\times p \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times q \times$

$q \times q \times q \times q \times q \times q \times q \times q$ .

Express the following number in standard form:
$0.00000000000942$

2124178_1671204_ans_22f1f76dd4bf4ec4bbe9905191e67eaa.jpg

Express the following Number in the standard form:
$846 \times 10^7$

2126809_1676660_ans_2a2955078ad64291944c7f6c55344d91.jpg

find the product of
$a^3\times 3ab^2\times 2a^2b^2$ .

We know that if a number say

$b$ is multiplied three times. That is,

$b\times b\times b$ can be written as

$b^3$ .

$b^3$ =

$b\times b\times b$

Similarly,

$a^{3} = a\times a \times a$ .

$3ab^{2} = 3\times a\times b\times b$ .

$2a^{2}b^{2} = 2 \times a \times a \times b \times b$ .

Thus,

$a^{3} \times 3 ab^{2}\times 2a^{2}b^{2} = a\times a \times a \times 3\times a\times b\times b \times 2 \times a \times a \times b \times b$ .

$= 6 \times a\times a \times a \times a\times b\times b \times a \times a \times b \times b$ .

$= 6 \times a\times a \times a \times a\times a \times a \times b \times b \times b\times b$ .

$= 6 \times a^{6} \times b^{4}$

$= 6 a^{6}b^{4}$ .

Express the number $5,00,00,000$ in the standard from.

1517375_1676654_ans_c18af6997fcc4e5cbda50a6db554d66e.jpg

Write the given statement in the product form.
$x^3y^4$ .

We know that,

$b^{3}$ =

$b\times b\times b$ .

Similarly,

$x^{3}= x\times x\times x$ .

$y^{4} = y \times y\times y\times y$ .

Thus,

$x^{3}y^{4}= x\times x\times x \times y \times y\times y\times y$ .

Write the following number in the usual form:
$4.83 \times 10^7$

Now,

$4.83\times 10^7$

$=\dfrac{483}{10^2}\times 10^7$

$=483\times 10^{(7-2)}$

$=483\times 10^5$

$=4,83,00,000$

Express the following Numbers in the standard from: $723 \times 10^9$

1518033_1676661_ans_5c44db87ccd540b69b813e1345778905.jpeg

The area of a rectangle is given by the product of its length and breadth. The length of a rectangle is two-third of its breadth. Find its area, if its breadth is x cm.

Let,

The length of the rectangle =

$l$ .

The breadth of the rectangle =

$b$ .

Given,

The breadth of a rectangle (

$b$ ) =

$x$ ---- (1)

The length of the rectangle (

$l$ )=

$\dfrac{2}{3}$ of breadth of rectangle.

$l$ =

$\dfrac{2}{3} \times b$ --- (2)

Area of a rectangle = Product of length and breadth

$l \times b$ .

$\dfrac{2}{3} \times b \times b$ . [From (2)]

$\dfrac{2}{3} x^{2}$ . [From (1)]

1409231_1674315_ans_eab502b3f28e4aaaa603a1cf1eea0b5e.png

Simplify $\left(\dfrac{81}{49}\right)^{-\tfrac{3}{2}}$ .

1680419_1715594_ans_fbec276dc041422bab362920a1dd2723.jpg

Write the following in ascending order of magnitude.
$\sqrt[6]{6}, \sqrt[3]{7}, \sqrt[4]{8}$ .

1886511_1715630_ans_6e627ebfbc73456d9f99b4eddd7abba6.jpg

Evaluate $8^{\tfrac{-1}{3}}$ .

1680413_1715588_ans_9f3d599e3ea34e54a7d7eb8c5b4d7e86.jpg

If $a=2, b=3$ find the value of $(a^b+b^a)^{-1}$ .

1680414_1715591_ans_79a781da72a342b38617dfa4ae6f1c4a.jpg

Simplify $\left(\dfrac{7776}{243}\right)^{-\dfrac{3}{5}}$ .

1680456_1715598_ans_1df053908a894225807522960a3e8f77.jpg

Evaluate:
$\left(\dfrac{1}{2}\right)^{-5}$

Given

$\left(\dfrac{1}{2}\right)^{-5}$

According to law of exponents, we can write

$\left(\dfrac{1}{2}\right)^{-5}$ as

$\left(\dfrac{1}{2}\right)^{-5}=2^{5}\left[\because a^{-m}=\left(\dfrac{1}{a^{m}}\right)\right]$

But we know that

$2^{5}$ is

$32, \left(\dfrac{1}{2}\right)^{-5}=32$

Evaluate:
$4^{-3}$

Given:

$4^{-3}$

According to law of exponents we can write

$4^{-3}$

$4^{-3}=\left(\dfrac{1}{4^{3}}\right) \left[\because a^{-m}=\left(\dfrac{1}{a^{m}}\right)\right]$

But we know that

$4^{3}$ is

$64$

Hence

$4^{-3}=\dfrac{1}{64}$

Evaluate:
$\left(\dfrac{4}{3}\right)^{-3}$

Given:

$\left(\dfrac{4}{3}\right)^{-3}$

According to law of exponents, we can write

$\left(\dfrac{4}{3}\right)^{-3}$ as

$\left(\dfrac{4}{3}\right)^{-3}=\left(\dfrac{3}{4}\right)^{3}$

$\left[\because \left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}\right]$

But we know that

$3^{3}$ is

$27$ and

$4^{3}$ is

$64$ , now substitute this in above

We get

$\left(\dfrac{3}{4}\right)^{3}=\dfrac{3^{3}}{4^{3}}$

$\therefore \dfrac{3^{3}}{4^{3}}=\dfrac{27}{64}$

Simplify $\left(\dfrac{32}{243}\right)^{-\tfrac{4}{5}}$ .

1680452_1715597_ans_03464b450a1d4c9892a1bbc6d927a85d.jpg

If $a=2, b=3$ find the value of $(a^a+b^b)^{-1}$ .

1680417_1715592_ans_8eaf5d3b5bb5464387e3857d6b069b19.jpg

Evaluate:
$\left\{\left(\dfrac{3}{2}\right)^{-2}\right\}^{2}$

Given that

$\left\{\left(\dfrac{3}{2}\right)^{-2}\right\}^{2}$

According to third law of exponents i.e.,

$\left\{\left(\dfrac{a}{b}\right)^{m}\right\}^{n}=\left(\dfrac{a}{b}\right)^{m\times n}$

We can apply this law in above equation we get

$\left\{\left(\dfrac{3}{2}\right)^{-2}\right\}^{2}=\left(\dfrac{3}{2}\right)^{-4}$

Again apply the third law of exponents,

Now

$\left(\dfrac{3}{2}\right)^{-4}=\left(\dfrac{2}{3}\right)^{4}=\dfrac{16}{81}$

$\left[\because \left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}\right]$

We know that

$3^{4}$ is

$81$ and

$2^{4}$ is

$16$

On substituting we get

$\left(\dfrac{2}{3}\right)^{4}=\dfrac{16}{81}$

Evaluate $\left\{\left(\dfrac{1}{3}\right)^{-3}-\left(\dfrac{1}{2}\right)^{-3}\right\}\div \left(\dfrac{1}{4}\right)^{-3}$

Given that

$\left\{\left(\dfrac{1}{3}\right)^{-3}-\left(\dfrac{1}{2}\right)^{-3}\right\}\div \left(\dfrac{1}{4}\right)^{-3}$

The above equation is in the form of

$\left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}$

By applying the law we get

$\left\{\left(\dfrac{1}{3}\right)^{-3}-\left(\dfrac{1}{2}\right)^{-3}\right\}\div \left(\dfrac{1}{4}\right)^{-3}= [3^{3}-2^{3}]\div (4^{3})$

On simplifying above equation by substituting the

$2^{3}$ is

$8,3^{3}$ is

$27$ and

$4^{3}$ is

$64$

We get

$\left\{3^{2}-2^{3}\right\}\div (4^{3})=(27-8)\div 64=\dfrac{19}{64}$

Evaluate : $\left (1 \dfrac{61} {64}\right )^{-\tfrac{2} {3}}$

Here we have

$\left (1 \dfrac{61} {64}\right )^{-\tfrac{2} {3}} = \left ( \dfrac{125} {64}\right )^{-\tfrac{2} {3}}\\ = \left ( \dfrac{5^{3}} {4^{3}}\right )^{-\tfrac{2} {3}}\\ = \left (\dfrac 54 \right )^{3 \times -\tfrac23}\\ = \left (\dfrac 54 \right )^{-2}\\ = \left (\dfrac45 \right )^{2}\\ =\dfrac{ 16}{25}$

Evaluate:
$\left(\dfrac{2}{3}\right)^{-3}\times \left(\dfrac{2}{3}\right)^{-2}$

Given

$\left(\dfrac{2}{3}\right)^{-3}\times \left(\dfrac{2}{3}\right)^{-2}$

According to first law of exponents,

$\left(\dfrac{a}{b}\right)^{m}\times \left(\dfrac{a}{b}\right)^{n}=\left(\dfrac{a}{b}\right)^{m+n}$

We can write above equation as

$\left(\dfrac{2}{3}\right)^{-3}\times \left(\dfrac{2}{3}\right)^{-2}=\left(\dfrac{2}{3}\right)^{-3+(-2)}= \left(\dfrac{2}{3}\right)^{-5}$

Now we have

$\left(\dfrac{2}{3}\right)^{-5}$ , this can be written in the form of

$a^{m}=\left(\dfrac{1}{a^{m}}\right)$

Then above equation becomes,

$\left(\dfrac{2}{3}\right)^{-5}= \left(\dfrac{3}{2}\right)^{5}$

But we know that

$3^{5}$ is

$243$ and the value of

$2^{5}$ is

$32$ ,

Substituting these in above equation, we get

$\left(\dfrac{3}{2}\right)^{5}=\left(\dfrac{243}{32}\right)$

Evaluate:
$\left\{\left(\dfrac{-2}{3}\right)^{2}\right\}^{-2}$

Given

$\left\{\left(\dfrac{-2}{3}\right)^{2}\right\}^{-2}$

According to third law of exponents i.e.

$\left\{\left(\dfrac{a}{b}\right)^{m}\right\}^{n}=\left(\dfrac{a}{b}\right)^{m\times n}$

We can apply this law in above equation we get

$\left\{\left(\dfrac{-2}{3}\right)^{2}\right\}^{-2}=\left(\dfrac{-2}{3}\right)^{2\times (-2)}$

Again apply the third law of exponents we get,

$\left(\dfrac{-2}{3}\right)^{2\times (-2)}=\left(\dfrac{-2}{3}\right)^{-4}$

Now above equation is in the form of

$\left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}$

So we get

$\left(\dfrac{-2}{3}\right)^{-4}=\left(\dfrac{3}{-2}\right)^{4}$

But we know that

$3^{4}$ is

$81$ and

$2^{4}$ is

$16$

$\therefore \left(\dfrac{3}{-2}\right)^{4}=\dfrac{81}{16}$

Evaluate:
$\left(\dfrac{-2}{3}\right)^{-5}$

Given

$\left(\dfrac{-2}{3}\right)^{-5}$

According to law of exponents,we can write

$\left(\dfrac{-2}{3}\right)^{-5}$ as

$\left(\dfrac{-2}{3}\right)^{-5}=\left(\dfrac{3}{-2}\right)^{5}$

$\left[\because \left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}\right]$

But we know that

$3^{5}$ is

$243$ and the value of

$2^{o}$ is

$32$

Now substituting these value in above equation,

We get

$\left(\dfrac{3}{-2}\right)^{5}=\left(\dfrac{3^{5}}{-2^{5}}\right)=\left(\dfrac{243}{-32}\right)$

$\therefore \left(\dfrac{3}{-2}\right)^{5}=-\dfrac{243}{32}$

Evaluate:
$\left(\dfrac{-2}{3}\right)^{-3}\times \left(\dfrac{-2}{3}\right)^{-2}$

Given

$\left(\dfrac{-2}{3}\right)^{-3}\times \left(\dfrac{-2}{3}\right)^{-2}$

According to

$a^{-m}=\left(\dfrac{1}{a^{m}}\right)$ we can write above equation as

$\left(\dfrac{-2}{3}\right)^{-3}\times \left(\dfrac{-2}{3}\right)^{-2}=\left(\dfrac{3}{-2}\right)^{3}\times \left(\dfrac{3}{-2}\right)^{2}$

Now apply the first law of exponents i.e.

$\left(\dfrac{a}{b}\right)^{m}\times \left(\dfrac{a}{b}\right)^{n}=\left(\dfrac{a}{b}\right)^{m+n}$

[

$\because$ Base in same in above equation]

We get

$\left(\dfrac{3}{-2}\right)^{3}\times \left(\dfrac{3}{-2}\right)^{-2}=\left(\dfrac{3}{-2}\right)^{3+2}=\left(\dfrac{3}{-2}\right)^{5}$

But we know that

$3^{5}$ is

$243$ and

$2^{5}$ is

$32$

$\therefore \left(\dfrac{3}{-2}\right)^{5}=\dfrac{243}{-32}$

Evaluate:
$\left[\left\{\left(\dfrac{-1}{3}\right)^{2}\right\}^{-2}\right]^{-1}$

Given

$\left[\left\{\left(\dfrac{-1}{3}\right)^{2}\right\}^{-2}\right]^{-1}$

According to third law of exponents i.e.

$\left\{\left(\dfrac{a}{b}\right)^{m}\right\}^{n}=\left(\dfrac{a}{b}\right)^{m\times n}$

We can apply this law in above equation we get

$\left\{\left(\dfrac{-2}{3}\right)^{2}\right\}^{-2}=\left(\dfrac{-2}{3}\right)^{2\times (-2)}$

Again apply the third law of exponents,

Now,

$\left\{\left(\dfrac{-1}{3}\right)^{2}\right\}^{2}=\left(\dfrac{-1}{3}\right)^{4}$

But we know that

$3^{4}=81$ , on substituting We get

$\left(\dfrac{-1}{3}\right)^{4}=\dfrac{1}{81}$

Evaluate $[\{(5^{-1})\times (3^{-1})\}^{-1}\div 6^{-1}]$

Given

$[\{(5^{-1})\times (3^{-1})\}^{-1}\div 6^{-1}]$

The above equation is in the form of

$\left(\dfrac{a}{b}\right)^{-m}=\left(\dfrac{b}{a}\right)^{m}$

By applying the law we get

$[\{(5^{-1})\times (3^{-1})\}^{-1}\div 6^{-1}]=\left[\left\{\left(\dfrac{1}{5}\right)\times \left(\dfrac{1}{3}\right)\right\}^{-1}\div \left(\dfrac{1}{6}\right)\right]$

On simplifying the above equation, we get:

$\left[\left\{\left(\dfrac{1}{5}\right)\times \left(\dfrac{1}{3}\right)\right\}^{-1}\div \left(\dfrac{1}{6}\right)\right]=\left[\left(\dfrac{1}{15}\right)^{-1}\div \left(\dfrac{1}{6}\right)\right]=\left(15\div \left(\dfrac{1}{6}\right)\right)$

$\Rightarrow \left(15 \times \dfrac{6}{1}\right)=90\quad \quad \left[\because \ \dfrac{a}{b} \div \dfrac{c}{d}=\dfrac{a}{b} \times \dfrac{d}{c}\right]$

Hence,

$[\{(5^{-1})\times (3^{-1})\}^{-1}\div 6^{-1}]=90$ .

Find the value of:
$\left(\dfrac{1}{2}\right)^{-1}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{4}\right)^{-2}$

Given that

$\left(\dfrac{1}{2}\right)^{-1}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{4}\right)^{-2}$

The above equation is in the form of

$a^{-m}=\left(\dfrac{1}{a^{m}}\right)$

By applying this law, we get:

$\left(\dfrac{1}{2}\right)^{-1}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{4}\right)^{-2}=2+3^{2}+4^{2}$

On simplifying, we get:

$2+3^{2}+4^{2}=2+9+16$

$=27$

Hence,

$\left(\dfrac{1}{2}\right)^{-1}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{4}\right)^{-2}=27$ .

Solve the following:
$100^{-10}$

Use

$x^{-m}=\dfrac{1}{x^m}$

$=100^{-10}=\dfrac{1}{100^{10}}$

Simplify :
$\left( \dfrac{1}{4}\right)^{-2}+\left( \dfrac 12\right)^{-2}+\left( \dfrac 13\right)^{-2}$

Use,

$a^{m}=\left(\dfrac{1}{a}\right)^{-m}$

$\therefore \left( \dfrac {1}{4}\right)^{-2}+\left( \dfrac {1}{2}\right)^{-2}+\left( \dfrac {1}{3}\right)^{-2}=(4)^2+(2)^2+(3)^2=16+4+9=29$

Simplify :
$\left[ \left( \dfrac{-2}{3}\right)^{-2}\right]^{3}\times \left( \dfrac{1}{3}\right)^{-4}\times 3^{-1}\times \dfrac 16$

Use,

$(a^m)^n=(a)^{mn}, a^{-m}=\dfrac{1}{a^m}, a^m \times a^n =a^{m+n}$ and

$a^m \div a^n =(a)^{m-n}$
Therefore,

$\left( \left( \dfrac{-2}{3}\right)^{-2}\right)^3\times \left( \dfrac{1}{3}\right)^{-4}\times 3^{-1}\times \dfrac 16=\left( \dfrac{-2}{3}\right)^{(-2)\times 3}\times (3)^4\times \dfrac 13 \times \dfrac 16$

$=\left( \dfrac {-2}{3}\right)^6 \times 3^4 \times \dfrac{1}{3}\times \dfrac{1}{2\times 3}$

$=\left( \dfrac{3}{-2}\right)^6 \times 3^4 \times \dfrac{1}{3\times 2\times 3}=\dfrac{(3)^6}{(2)^6}\times 3^4 \times \dfrac{1}{2^1 \times 3^2}$

$=\dfrac{(3)^{6+4}}{(2)^{6+1}\times 3^2}$

$=\dfrac{(3)^{10}}{2^7 \times 3^2}=\dfrac{3^{10-2}}{2^7}=\dfrac{3^8}{2^7}$

Simplify :
$\dfrac{49\times z^{-3}}{7^{-3}\times 10\times z^{-5}}( z\neq 0)$

Use

$a^m \div a^n =(a)^{m-n}$

$\Rightarrow \dfrac{49\times z^{-3}}{7^{-3}\times 10\times z^{-5}}=\dfrac{(7)^2\times z^{-3}}{7^{-3}\times 10\times z^{-5}}=\dfrac{(7)^{2+3 }\times z^{-3+5}}{10}$

$\Rightarrow \dfrac{49\times z^{-3}}{7^{-3}\times 10\times z^{-5}}=\dfrac{(7)^5 \times z^2}{10}=\dfrac{7^5}{10}z^2$

Express as a power of a rational number with negative exponent.
$\left[ \left( \dfrac{-3}{2}\right)^{-2}\right]^{-3}$

Use

$(a^m)^n =(a)^{mn}$ and

$a^{-m}=\dfrac{1}{a^m}$
We have,

$\left( \left( \dfrac{-3}{2}\right)^{-2}\right)^{-3}=\left( \dfrac{-3}{2}\right)^{-2\times (-3)}=\left( \dfrac{-3}{2}\right)^6=\left( \dfrac{3}{2}\right)^{6}$

$\therefore \left( \left( \dfrac{-3}{2}\right)^{-2}\right)^{-3}=\left( \dfrac{3}{2}\right)^{6}$

Find the value of $[4^{-1}+3^{-1}+6^{-2}]^{-1}$

$x^{-x}=\dfrac{1}{x^m}$

$\Rightarrow [4^{-1}+3^{-1}+6^{-2}]^{-1}$

${ \left[ \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 3 } +\dfrac { 1 }{ { 6 }^{ 2 } } \right] }^{ -1 }={ \left[ \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 36 } \right] }^{ -1 }$

$\Rightarrow [4^{-1}+3^{-1}+6^{-2}]^{-1} == { \left[ \dfrac { 9+12+1 }{ 36 } \right] }^{ -1 }$

$\Rightarrow [4^{-1}+3^{-1}+6^{-2}]^{-1} ={ \left[ \dfrac { 22 }{ 36 } \right] }^{ -1 }$

$\Rightarrow [4^{-1}+3^{-1}+6^{-2}]^{-1} ={ \left[ \dfrac { 36 }{ 22 } \right] }= \dfrac{18}{11}$

If $36=6 \times 6=6^2$ , then $\dfrac{1}{36}$ expressed as a power with base $6$ is ______.

$x^{-m}=\dfrac{1}{x^m}$

$\Rightarrow 36=6 \times 6=6^2$

$\therefore \dfrac{1}{36} =\dfrac{1}{6 \times 6}$

$\Rightarrow \dfrac{1}{36} = \dfrac{1}{6 \times 6}$

$\Rightarrow \dfrac{1}{36} = \dfrac{1}{6 ^2}$

$\Rightarrow \dfrac{1}{36} = 6^{-2}$

The value of $[3^{-1} \times \ 4^{-1}]^2$ is _______.

$[3^{-1} \times \ 4^{-1}]^2 =\dfrac{1}{(3 \times 4)^2} =\dfrac{1}{(12)^2} =\dfrac{1}{(4 \times 3)^2} =\dfrac{1}{(4^2 \times 3)^2}$

If $\dfrac{5^m \times 5^3 \times 5^{-2}}{5^{-5}}=5^{12}$ . Find $m$ .

Using, laws of exponents,

$a^m \div a^n =(a)^{m-n}$ and

$a^{-m}=\dfrac{1}{a^m}$

Then, equation becomes

$5^m \times 5^3 \times 5^{-2}\times 5^5=5^{12}$

Now using,

$a^m \times a^n =a^{m+n}$

$\Rightarrow 5^m \times 5^8 \times 5^{-2}=5^{12}$

$\Rightarrow 5^m \times 5^6=5^{12}$

$\Rightarrow 5^{m+6}=5^{12}$

Comparing both sides

$\Rightarrow m+6=12$

$\Rightarrow m=6$

Evaluate
$(2/7)^{-4}$

Let us evaluate the given expression,

By using the formula

$(a^{-n}=1/a^n)$

So,

$(2/7)^{-4}=(7/2)^4$

$=(7/2)\times (7/2)\times (7/2)\times (7/2)$

$=2401/16$

Consider a quantity of a radioactive substance. The fraction of this quantity that remains after $t$ half - lives can be found by using the expression $3^{-t}$ .
What fraction of substance remains after $7$ half-lives?

Since,

$3^{-t}$ is the expression used for finding the fraction of the quantity that remains after

$t$ half-lives.

Hence, the fraction of substance remains after

$7$ half-lives will be equal to

$3^{-7}$ i.e.

$\dfrac{1}{3^7}$ .

In a shrinking machine, a piece of stick is compressed to reduce its length. If $9\ cm$ long sandwich is put into the shrinking machine below, how many $cm$ long will it be when it emerges?

In given machine

$\cfrac13$ is base because

$(\times \cfrac13)$ is given and

$-1$ is power of base.

So, machine will shrink the object by

$\left(\times\left(\cfrac13\right)^{-1}=\times3\right)$

So, if we put a

$9\ cm$ long sandwich into the shrinking machine then the length of sandwich would become

$=9\times \dfrac{1}{3^{-1}}=9\times 3=27\ cm$

What happens when $1\ cm$ worms are sent through these hook-ups?

$a$ is input to

$\times x^y$ machine then output will be

$a\times x^y$

When two such machines are hooked up, output of one is input to other.

So, for overall output their stretching power multiplies.

So,

If you put

$1\ cm$ worms through

$( \times 2^1 )$ and

$(\times 2^{-1})$ machine,

the result will be

$=1\times 2\times2^{-1}=1\times 2\times \dfrac 12=1\ cm$ .

That is worm will remain as it is, after going through both machines.

Find the value of
$\dfrac{6^n}{6^{-2}}=6^3$

Given,

$\dfrac{6^n}{6^{-2}}=6^3$

$a^m \div a^n=(a)^{m-n}$
Using law of exponent

$\Rightarrow 6^{n-(-2)}=6^3$
On comparing both sides-

$n+2=3$

$n=1$
Therefore the value of

$n$ is

$1$ .

Simplify
$[(2)^{-1}+(4)^{-1}+(3)^{-1}]^{-1}$

Let us simplify the given expression

$[(2)^{-1}+(4)^{-1}+(3)^{-1}]^{-1}=[(1/2)+(1/4)+(1/3)]^{-1}$

$=[(6+3+4)/12]^{-1}$

$=[13/12]^{-1}$

$=12/13$

Evaluate:
$(0.25)^{-1}$ .

$(0.25)^{-1}=\dfrac1{0.25}=\dfrac{100}{25}=4$

Evaluate:
$5^{-4} \times (125)^{\dfrac {5}{3}}- (25)^{-\dfrac 12}$

$5^{-4} \times (125)^{\dfrac {5}{3}}- (25)^{-\dfrac 12}=5^{-4}\times (5\times 5\times 5)^{\dfrac 53}\div (5\times 5)^{-\dfrac 12}$

$=5^{-4}\times (5^{3})^{\dfrac 53}\div (5^2)^{-\dfrac 12}$

$=5^{-4}\times \left(5^{3+\dfrac 53}\right) \div \left(5^{2\left(-\dfrac 12 \right)}\right)$

$=\dfrac {5^{-4}\times 5^{3}}{5^{-1}}$

$=\dfrac {5^{5-4}}{5^{-1}}$

$=\dfrac {5^{-1}}{5^{-1}}$

$=5^{1-(-1)}$

$=5^{2}$

$=5\times 5$

$=25$

Simplify:
$(a+b)^{-1}, (a^{-1}+b^{-1})$

$(a+b)^{-1}, (a^{-1}+b^{-1})=\dfrac {1}{(a+b)}\times \left(\dfrac {1}{a}+\dfrac {1}{b}\right)$

$=\dfrac {1}{a+b}\times \left(\dfrac {b+a}{ab}\right)$

$=\dfrac {1}{a+b}\times \dfrac {(a+b)}{ab}$

$=\dfrac {1}{ab}$

Evaluate : $\left (0.027 \right )^{-1/3}$

Here we have,

$\ \left (0.027\right )^{-1/3}\\ = \left (\dfrac{27}{1000 }\right )^{-1/3}\\ = \left [ \dfrac{\left (3 \times 3 \times 3 \right )}{ \left (10 \times 10 \times 10 \right )} \right ]^{-1/3}\\ = \left (\dfrac{3^{3}}{10^{3} } \right )^{-1/3}\\ = \left (\dfrac 3{10} \right )^{3 \times -\tfrac13}\\ = \left (\dfrac 3{10} \right )^{-1}\\ = \dfrac{10}3$

$(m+n)^{-1}(m^{-1}+n^{-1}) =(mn)^{-1}$

L.H.S

$(m+n)^{-1}(m^{-1}+n^{-1})$

$=\dfrac{1}{m+n}(\dfrac{1}{m}+\dfrac{1}{n})\\=\dfrac{1}{m+n}\times\dfrac{n+m}{mn}=\dfrac{1}{mn}$

$=(mn)^{-1}$
=R.H.S.
Hence proved.

Evaluate: $\left ( \dfrac{81}{16} \right )^{-3/4}$

Here we have,

$\ \ \left ( \dfrac{81}{16} \right )^{-3/4} = \left ( \dfrac{3^{4}}{2^{4} }\right ) ^{-3/4}$

$\quad = \left [\left ( \dfrac 32 \right )^{4} \right ]^{-3/4}\\ \quad = \left ( \dfrac 32 \right )^{(-3/4) \times 4}$

$\quad= \left ( \dfrac 32 \right )^{-3} \\ \quad= \left ( \dfrac23 \right )^{3}\\ \quad= \dfrac{2^{3}}{3^{3}} \\ \quad= \dfrac{\left (2 \times 2 \times 2 \right )}{\left ( 3 \times 3 \times 3 \right )}\\ \quad=\dfrac 8{27}$

$\left (1\frac{61} {64} \right )^{-\frac{2} {3}}$

$\left (1\frac{61} {64} \right )^{-\frac{2} {3}} = \left (\frac{125} {64} \right )^{-\frac{2} {3}} = \left (\frac{5^{3}} {4^{3}} \right )^{-\frac{2} {3}}$

$= \left ( 5/4 \right )^{3 \times -2/3}$

$= \left ( 5/4 \right )^{-2}$

$= \left (4/5 \right )^{2}$

$= 16/25$

Prove that :
$\dfrac{a^{-1}}{a^{-1}+b^{-1}}+\dfrac{a^{-1}}{a^{-1}-b^{-1}}=\dfrac{2b^2}{b^2-a^2}$

$\dfrac{a^{-1}}{a^{-1}+b^{-1}}+\dfrac{a^{-1}}{a^{-1}-b^{-1}}=\dfrac{2b^2}{b^2-a^2}$

$L.H.S=\dfrac{a^{-1}}{a^{-1}+b^{-1}}+\dfrac{a^{-1}}{a^{-1}-b^{-1}}$

$=\dfrac{\dfrac{1}{a}}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{\dfrac{1}{a}}{\dfrac{1}{a}-\dfrac{1}{b}}$

$=\dfrac {\dfrac{1}{a}}{\dfrac{b+a}{ab}}+\dfrac{\dfrac{1}{a}}{\dfrac{b-a}{ab}}$

$=\dfrac{1}{a}\times \dfrac{ab}{b+a}+\dfrac{1}{a}\times \dfrac{ab}{b-a}$

$=\dfrac{b}{b+a}+\dfrac{b}{b-a}$

$=\dfrac{b^2-ab+b^2+ab}{b^2-a^2}$

$=\dfrac{2b^2}{b^2-a^2}$

$=R.H.S.$

Simplify each of the following and express with positive index:
$[1-(1-(1-n)^{-1})^{-1}]^{-1}$

$[1-(1-(1-n)^{-1})^{-1}]^{-1} =\dfrac {1}{[1-(1-(1-n)^{-1})^{-1}]^{-1}}$

$=\dfrac {1}{1-\dfrac {1}{1-(1-n)^{-1}}}$

$=\dfrac {1}{1-\dfrac {1}{(1-n)}}$

$=\dfrac {1}{1-\dfrac {1}{\dfrac {1(1-n)-1}{(1-n)}}}$

$=\dfrac {1}{1-\dfrac {1}{\dfrac {1-n-1}{(1-n)}}}$

$=\dfrac {1}{1-\dfrac {1}{\dfrac {-n}{(1-n)}}}$

$=\dfrac {1}{1-\dfrac {(1-n)}{-n}}$

$=\dfrac {1}{1+\dfrac {(1-n)}n{}}$

$=\dfrac {1}{\dfrac {n+(1-n)}{n}}$

$=\dfrac {1}{\dfrac {n+1-n}{n}}$

$=\dfrac {n}{1}$

${ \left( \frac { 1 }{ 2 } \right) }^{ -2 }+{ \left( \frac { 1 }{ 3 } \right) }^{ -2 }+{ \left( \frac { 1 }{ 4 } \right) }^{ -2 }=?$

2040779_1327564_ans_8b316d0538f5408993659280aa7bcd05.jpg

Express the number appearing in the following statement in standard from.
Diameter of the sun is $1,400,000,000 \text{ m}$

Diameter of sun

$=1,400,000,000 \text{ m}$

In standard form,

$\Rightarrow$ Diameter of sun

$=\left( 1.4\times 10^9\right) \text{ m}$

By what a number should $\left (\dfrac {5}{3}\right)^{-2}$ be multiplied so that the product may be equal to $\left (\dfrac {7}{3}\right)^{-1}$ ?

2039750_1671150_ans_014d188d26484118b43d6b2f5db1573d.jpg

By what a number should $\left (\dfrac {1}{2}\right)^{-1}$ be multiplied so that the product may be equal to $\left (\dfrac {-4}{7}\right)^{-1}$ ?

2039749_1671144_ans_34dce122f7d944c2867033ca8780e8b5.jpg

Express the number appearing in the following statement in standard from.
Mass of Uranus is $86,800,000,000,000,000,000,000,000\ kg$

2122819_1218733_ans_cc6f8801a2da4b93ae665a033d1f7530.jpg

The value of $5\sqrt[4]{\sqrt[3]{5}}-\sqrt[3]{\sqrt[4]{5}}$ is?

1328855_1587682_ans_2aacf8a670ef41dbba41c0e4ed3a0c8c.jpg

Exponents And Powers - Class 8 Maths - Extra Questions

Evaluate : (8125)−13\left (\dfrac 8{125} \right )^{-\tfrac13}

(−127)−2/3\left ( \dfrac{-1}{27} \right )^{-2/3}

Identify the greater number, wherever possible, in the following.2102^{10} or 10210^{2}.

Express the number appearing in the following statement in standard form.The earth has 1,535,000,0001,535,000,000 cubic kmkm of seawater.

Express the number appearing in the following statement in standard form.Diameter of the Earth is 1,27,56,0001,27,56,000 meters.

Express the the mean distance between Earth and moon is 384,000,000 m384,000,000\ m in standard from.

Express the number appearing in the following statement in standard from.The univers is estimated to be 12,000,000,00012,000,000,000 years old.

Express the number appearing in the following statement in standard from:The distance between Sun and Saturn is 1,433,000,000,000 m .1,433,000,000,000\ m\ .

Write the given product in exponential form.7×a×a×a.....87\times a\times a\times a..... 8 times ×b×b×b×…5\times b\times b\times b\times … 5 times.

Write the given expression in the product form.30x4y4z530x^4y^4z^5

Write the given product in exponential form.4×a×a×…..5 times ×b×b×…..12 times ×c×c.....15 times 4\times a\times a\times ….. 5\text{ times }\times b\times b\times ….. 12\text{ times }\times c\times c..... 15\text{ times }.

Express the numbers appearing in the following statements in the standard form:The universe is estimated to be about 12,000,000,000 years old.

Express the numbers appearing in the following statements in the standard form:Diameter of the Earth is 1,27,56,000 metres.

Evaluate 64−1264^{\tfrac{-1}{2}}.

Identify the greater number, wherever possible, in the following?1002100^2 or 21002^{100}.

Identify the greater number in the following.282^8 or 828^2.

Write the following number in standard form:11000000\dfrac{1}{1000000}.

Evaluate(35)−2\left(\dfrac 3 5\right)^{-2}

Express the number appearing in the following statement in standard from.60,230,000,000,000,000,000,00060,230,000,000,000,000,000,000 molecules are present in a drop of water weighing 1.8 gm1.8\ gm.

Which is greater?(12)12\left (\dfrac{1}{2} \right )^{\frac{1}{2}} or (23)13\left (\dfrac{2}{3} \right )^{\frac{1} {3}}

Prove that:(√3×5−3÷3√3−1√5)×6√3×56=35(\sqrt{3 \times 5^{-3}} \div \sqrt [ 3 ]{ 3^{ -1} } \sqrt{5}) \times \sqrt [ 6 ]{ 3 \times 5^6 } = \dfrac{3}{5}

Express 16−216^{-2} as a power with base 44

Which is greater 7927^{92} or 8918^{91}?

Which is larger: (100)4{ \left( 100 \right) }^{ 4 } or (125)3 { \left( 125 \right) }^{ 3 }?

Simplify and give reasons(34)−3\left(\dfrac{3}{4}\right)^{-3}

Express the following in standard from:i) 76854000×10876854000\times 10^{8}ii) 0.000089×1060.000089\times 10^{6}

Simplify (5−3)2×34(3−2)−3×(53)−2\displaystyle \frac { \left( 5 ^ { - 3 } \right) ^ { 2 } \times 3 ^ { 4 } } { \left( 3 ^ { - 2 } \right) ^ { - 3 } \times \left( 5 ^ { 3 } \right) ^ { - 2 } }

Compare the following numbers:2.7×1012;1.5×1082.7\times{10}^{12};1.5\times{10}^{8}

Simplify and write the answer in exponential form(−4)−3×(5)3×(−5)−3{ \left( -4 \right) }^{ -3 }\times { \left( 5 \right) }^{ 3 }\times { \left( -5 \right) }^{ -3 }

Evaluate {(43)−1−(14)−1}−1\left\{\left(\dfrac{4}{3}\right)^{-1}-\left(\dfrac{1}{4}\right)^{-1}\right\}^{-1}

Express 512512 in power notation.

Prove that:{(13)−1−(14)−1}−1=−1{ \left\{ { \left( \dfrac { 1 }{ 3 } \right) }^{ -1 }-{ \left( \dfrac { 1 }{ 4 } \right) }^{ -1 } \right\} }^{ -1 } = -1

Say true or false and justify your answer:23>52{2}^{3}> {5}^{2}

Evaluate:[(−23)−2]3×(13)−4×3−1×16\left[\left(-\dfrac{2}{3}\right)^{-2}\right]^3\times \left(\dfrac{1}{3}\right)^{-4}\times 3^{-1}\times \dfrac{1}{6}

Find:125−13125^{\frac{-1}{3}}

Evaluate(−3)−3(-3)^{-3}

Evaluate:(i) (13)−1−(14)−1{\left (\dfrac{1}{3}\right )^{-1} - \left (\dfrac{1}{4}\right )^{-1}} (ii) (58)−7×(85)−4\left (\dfrac{5}{8}\right )^{-7}\times \left (\dfrac{8}{5}\right )^{-4}

Which is greater (150)300(150)^{300} or (20000)100×(100)100(20000)^{100}\times (100)^{100}?

Evaluate:(i) 3−23^{-2}(ii) (−4)−2(-4)^{-2}(iii) (12)−5\left (\dfrac{1}{2}\right)^{-5}

Show that (1.1)10000>1000(1.1)^{10000} > 1000

Find xx in the expression (18)3.5÷(27)3.5×63.5=2x(18)^{3.5}\div (27)^{3.5}\times 6^{3.5}=2^{x}

Mass of the earth is 5.97×1024 kg5.97\times {10}^{24}\ kg and mass of the moon is 7.35×1022 kg7.35\times {10}^{22}\ kg. What is the difference of their masses?

Compare 7×10−67 \times 10^{-6} and 129×10−7.129 \times 10^{-7}.

Which one is greater between 311431^{14} and 171817^{18}?

Express the following in the decimal form:4.67×1044.67\times {10}^{4}

Simplify: (8116)−3/4×(259)−3/2×(25)−3\left(\dfrac{81}{16}\right)^{-3/4} \times \left(\dfrac{25}{9}\right)^{-3/2} \times \left(\dfrac{2}{5}\right)^{-3}

Find the value of xx for which{(−27)2}x×(−72)−1=−8343{ \left\{ { \left( \dfrac { -2 }{ 7 } \right) }^{ 2 } \right\} }^{ x }\times { \left( \dfrac { -7 }{ 2 } \right) }^{ -1 }=\dfrac { -8 }{ 343 }.

If 9x×32×(3(−x2))−2=1279^{x} \times 3^{2}\times \left(3^{\left(\tfrac{-x}{2}\right)}\right)^{-2}=\dfrac{1}{27}, then find the value of xx.

If (53)−15×(53)11=(53)8x\left(\dfrac{5}{3}\right)^{-15} \times \left(\dfrac{5}{3}\right)^{11} =\left(\dfrac{5}{3}\right)^{8x}, then xx =?

If the standard form of 12,000,000,00012,000,000,000 is 1.2×10m1.2\times 10^{m} then find the value of mm.

If (64)x=1(256)y=2√2{ \left( 64 \right) }^{ x }=\cfrac { 1 }{ { \left( 256 \right) }^{ y } } =2\sqrt { 2 } ; then show that 3x+4y=03x+4y=0.

If 2x−1+2x+1=12802^{x-1}+2^{x+1}=1280, then find the value of xx.

Identify the greater number, wherever possible, in the following?535^3 or 353^5.

If x1/p=y1/q=z1/rx^{1/p} = y^{1/q} = z^{1/r} and xyz=1,xyz = 1, then the value of p+q+r=?p+q+r = ?

Identify the greater number between the following:434^3 or 343^4

Which is greater among the following?√2,3√4\sqrt{2}, \sqrt[3]{4} and 4√3 \sqrt[4]{3}

Write the base and the exponent of (7x)2(7x)^{2}

Change the given number into it's standard form128000000128000000

(5√8)52×(16)−32(\sqrt [ 5 ]{ 8 } )^{\dfrac{5}{2}}\times (16)^{\dfrac{-3}{2}}

Express the following as a product of prime factors in exponential form : 729×64729\times 64

Identify the greater number in each of the following numbers:28{2}^{8} or 82{8}^{2}

Identify the greater number among the two numbers:210{2}^{10} or 102{10}^{2}

Identify the greater number in the following numbers:1002{100}^{2} or 2100{2}^{100}

Express the number appearing in the following statements in standard form:602,300,000,000,000,000,000,000602,300,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8gm1.8gm.

Write 629000629000 in standard form.

Express the number appearing in the following statement in standard form:The Earth has 1,353,000,0001,353,000,000 cubic km of seawater.

Identify the greater number, wherever possible, in each of the following:53{5}^{3} and 35{3}^{5}

On solving the equation 72x+3=1 7^{2x+3} = 1 , we get x=−3bx=\dfrac{-3}{b} then find the value of bb.

Standard from of 60200000000000006020000000000000 is 6.02×10166.02\times 10^{16}.Enter 1 if it is True or 0 if it is False

Express the following numbers in standard from:4÷1000004 \div 100000

Write the given expression in the product form.43p10q5r1543p^{10}q^5r^{15}

Write the given product in exponential form.8×b×b×b×a×a×a×a8\times b\times b\times b\times a\times a\times a\times a.

Write the expression 5xy×3x2y×7y25xy\times 3x^2y\times 7y^2 in exponential form.

Write the given expression in the product form.a2b5a^2b^5.

Write the given expression in the product form.17p12q2017p^{12}q^{20}.

Express the following number in standard form:0.000000000009420.00000000000942

Express the following Number in the standard form:846×107 846 \times 10^7

find the product ofa3×3ab2×2a2b2a^3\times 3ab^2\times 2a^2b^2.

Express the number 5,00,00,0005,00,00,000 in the standard from.

Evaluate : $\left (\dfrac 8{125} \right )^{-\tfrac13}$

$\left ( \dfrac{-1}{27} \right )^{-2/3}$

Identify the greater number, wherever possible, in the following.
$2^{10}$ or $10^{2}$ .

Express the number appearing in the following statement in standard form.
The earth has $1,535,000,000$ cubic $km$ of seawater.

Express the number appearing in the following statement in standard form.
Diameter of the Earth is $1,27,56,000$ meters.

Express the the mean distance between Earth and moon is $384,000,000\ m$ in standard from.

Express the number appearing in the following statement in standard from.
The univers is estimated to be $12,000,000,000$ years old.

Express the number appearing in the following statement in standard from:
The distance between Sun and Saturn is $1,433,000,000,000\ m\ .$

Write the given product in exponential form.
$7\times a\times a\times a..... 8$ times $\times b\times b\times b\times … 5$ times.

Write the given expression in the product form.
$30x^4y^4z^5$

Write the given product in exponential form.
$4\times a\times a\times ….. 5\text{ times }\times b\times b\times ….. 12\text{ times }\times c\times c..... 15\text{ times }$ .

Express the numbers appearing in the following statements in the standard form:
The universe is estimated to be about 12,000,000,000 years old.

Express the numbers appearing in the following statements in the standard form:
Diameter of the Earth is 1,27,56,000 metres.

Evaluate $64^{\tfrac{-1}{2}}$ .

Identify the greater number, wherever possible, in the following?
$100^2$ or $2^{100}$ .

Identify the greater number in the following.
$2^8$ or $8^2$ .

Write the following number in standard form:
$\dfrac{1}{1000000}$ .

Evaluate
$\left(\dfrac 3 5\right)^{-2}$

Express the number appearing in the following statement in standard from.
$60,230,000,000,000,000,000,000$ molecules are present in a drop of water weighing $1.8\ gm$ .

Which is greater?
$\left (\dfrac{1}{2} \right )^{\frac{1}{2}}$ or $\left (\dfrac{2}{3} \right )^{\frac{1} {3}}$

Prove that:
$(\sqrt{3 \times 5^{-3}} \div \sqrt [ 3 ]{ 3^{ -1} } \sqrt{5}) \times \sqrt [ 6 ]{ 3 \times 5^6 } = \dfrac{3}{5}$

Express $16^{-2}$ as a power with base $4$

Which is greater $7^{92}$ or $8^{91}$ ?

Which is larger: ${ \left( 100 \right) }^{ 4 }$ or ${ \left( 125 \right) }^{ 3 }$ ?

Simplify and give reasons
$\left(\dfrac{3}{4}\right)^{-3}$

Express the following in standard from:
i) $76854000\times 10^{8}$
ii) $0.000089\times 10^{6}$

Simplify $\displaystyle \frac { \left( 5 ^ { - 3 } \right) ^ { 2 } \times 3 ^ { 4 } } { \left( 3 ^ { - 2 } \right) ^ { - 3 } \times \left( 5 ^ { 3 } \right) ^ { - 2 } }$

Compare the following numbers:
$2.7\times{10}^{12};1.5\times{10}^{8}$

Simplify and write the answer in exponential form
${ \left( -4 \right) }^{ -3 }\times { \left( 5 \right) }^{ 3 }\times { \left( -5 \right) }^{ -3 }$

Evaluate $\left\{\left(\dfrac{4}{3}\right)^{-1}-\left(\dfrac{1}{4}\right)^{-1}\right\}^{-1}$

Express $512$ in power notation.

Prove that:
${ \left\{ { \left( \dfrac { 1 }{ 3 } \right) }^{ -1 }-{ \left( \dfrac { 1 }{ 4 } \right) }^{ -1 } \right\} }^{ -1 } = -1$

Say true or false and justify your answer:
${2}^{3}> {5}^{2}$

Evaluate:
$\left[\left(-\dfrac{2}{3}\right)^{-2}\right]^3\times \left(\dfrac{1}{3}\right)^{-4}\times 3^{-1}\times \dfrac{1}{6}$

Find:
$125^{\frac{-1}{3}}$

Evaluate
$(-3)^{-3}$

Evaluate:
(i) ${\left (\dfrac{1}{3}\right )^{-1} - \left (\dfrac{1}{4}\right )^{-1}}$
(ii) $\left (\dfrac{5}{8}\right )^{-7}\times \left (\dfrac{8}{5}\right )^{-4}$

Which is greater $(150)^{300}$ or $(20000)^{100}\times (100)^{100}$ ?

Evaluate:
(i) $3^{-2}$ ii) $(-4)^{-2}$ $\left (\dfrac{1}{2}\right)^{-5}$

Show that $(1.1)^{10000} > 1000$

Find $x$ in the expression $(18)^{3.5}\div (27)^{3.5}\times 6^{3.5}=2^{x}$

Mass of the earth is $5.97\times {10}^{24}\ kg$ and mass of the moon is $7.35\times {10}^{22}\ kg$ . What is the difference of their masses?

Compare $7 \times 10^{-6}$ and $129 \times 10^{-7}.$

Which one is greater between $31^{14}$ and $17^{18}$ ?

Express the following in the decimal form:
$4.67\times {10}^{4}$

Simplify: $\left(\dfrac{81}{16}\right)^{-3/4} \times \left(\dfrac{25}{9}\right)^{-3/2} \times \left(\dfrac{2}{5}\right)^{-3}$

Find the value of $x$ for which ${ \left\{ { \left( \dfrac { -2 }{ 7 } \right) }^{ 2 } \right\} }^{ x }\times { \left( \dfrac { -7 }{ 2 } \right) }^{ -1 }=\dfrac { -8 }{ 343 }$ .

If $9^{x} \times 3^{2}\times \left(3^{\left(\tfrac{-x}{2}\right)}\right)^{-2}=\dfrac{1}{27}$ , then find the value of $x$ .

If $\left(\dfrac{5}{3}\right)^{-15} \times \left(\dfrac{5}{3}\right)^{11} =\left(\dfrac{5}{3}\right)^{8x}$ , then $x$ =?

If the standard form of $12,000,000,000$ is $1.2\times 10^{m}$ then find the value of $m$ .

If ${ \left( 64 \right) }^{ x }=\cfrac { 1 }{ { \left( 256 \right) }^{ y } } =2\sqrt { 2 }$ ; then show that $3x+4y=0$ .

If $2^{x-1}+2^{x+1}=1280$ , then find the value of $x$ .

Identify the greater number, wherever possible, in the following?
$5^3$ or $3^5$ .

If $x^{1/p} = y^{1/q} = z^{1/r}$ and $xyz = 1,$ then the value of $p+q+r = ?$

Identify the greater number between the following:
$4^3$ or $3^4$

Which is greater among the following?
$\sqrt{2}, \sqrt[3]{4}$ and $\sqrt[4]{3}$

Write the base and the exponent of $(7x)^{2}$

Change the given number into it's standard form
$128000000$

$(\sqrt [ 5 ]{ 8 } )^{\dfrac{5}{2}}\times (16)^{\dfrac{-3}{2}}$

Express the following as a product of prime factors in exponential form :
$729\times 64$

Identify the greater number in each of the following numbers:
${2}^{8}$ or ${8}^{2}$

Identify the greater number among the two numbers:
${2}^{10}$ or ${10}^{2}$

Identify the greater number in the following numbers:
${100}^{2}$ or ${2}^{100}$

Express the number appearing in the following statements in standard form:
$602,300,000,000,000,000,000,000$ molecules are contained in a drop of water weighing $1.8gm$ .

Write $629000$ in standard form.

Express the number appearing in the following statement in standard form:
The Earth has $1,353,000,000$ cubic km of seawater.

Identify the greater number, wherever possible, in each of the following:
${5}^{3}$ and ${3}^{5}$

On solving the equation $7^{2x+3} = 1$ , we get $x=\dfrac{-3}{b}$ then find the value of $b$ .

Standard from of $6020000000000000$ is $6.02\times 10^{16}$ .
Enter 1 if it is True or 0 if it is False

Express the following numbers in standard from:
$4 \div 100000$

Write the given expression in the product form.
$43p^{10}q^5r^{15}$

Write the given product in exponential form.
$8\times b\times b\times b\times a\times a\times a\times a$ .

Write the expression $5xy\times 3x^2y\times 7y^2$ in exponential form.

Write the given expression in the product form.
$a^2b^5$ .

Write the given expression in the product form.
$17p^{12}q^{20}$ .

Express the following number in standard form:
$0.00000000000942$

Express the following Number in the standard form:
$846 \times 10^7$

find the product of
$a^3\times 3ab^2\times 2a^2b^2$ .

Express the number $5,00,00,000$ in the standard from.

Write the given statement in the product form.
$x^3y^4$ .