Match the following provided that $$a$$ and $$b$$ any two rational numbers different from zero and $$x, y$$ are any two rational numbers. $$(1)$$ $$a^{x} \times a^{y}$$ $$(a)$$ $$a^{x - y}$$ $$(2)$$ $$a^{x} \div a^{y}$$ $$(b)$$ $$a^{xy}$$ $$(3)$$ $$(a^{x})^{y}$$ $$(c)$$ $$a^{x + y}$$ $$(4)$$ $$(ab)^{x}$$ $$(d)$$ $$\dfrac {a^{x}}{b^{x}}$$ $$(5)$$ $$\left (\dfrac {a}{b}\right )^{x}$$ $$(e)$$ $$a^{x}\times b^{x}$$

$$1 - (c), 2 - (a) , 3 - (b), 4 - (e), 5 - (d)$$

$$1 - (b), 2 - (c) , 3 - (a), 4 - (e), 5 - (d)$$

$$1 - (a), 2 - (c) , 3 - (d), 4 - (e), 5 - (b)$$

$$1 - (a), 2 - (b) , 3 - (c), 4 - (d), 5 - (e)$$

MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 4

$p^{0}$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
$p$

Explanation

As we can see that

$p$ is raised to power

$0$ . It means that there is no term of

$p$ .

So,

$p^{0}=1$

Any real number when raised to the power

$0$ gives

$1$ .

$\log _8 m+{\log}_8\dfrac{1}{6}=\dfrac{2}{3}$ , then m is equal to

0%
24
0%
18
0%
12
0%
4

Explanation

$\log_8m+\log_8\cfrac16=\cfrac23$

$\Rightarrow \log_8\cfrac m6=\cfrac23$

$\Rightarrow \cfrac m6=8^{\cfrac23}$

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

$\Rightarrow m=24$

Hence, A is the correct option.

$\log {(2\times 3\times 4)}$ is equal to:

0%
$\log {2} + \log {3} +\log {4}$
0%
$\log {12}$
0%
$\log {9}$
0%
$\log {4}$

Explanation

$\log(2\times3\times4)$

$=\log2+\log3+\log4$

$(\log(a\times b)=\log a+\log b)$

Hence, A is the correct option.

Evaluate

$2^0+3^0$

0%
$2$
0%
$3$
0%
$5$
0%
$1$

Match the following provided that

$a$ and

$b$ any two rational numbers different from zero and

$x, y$ are any two rational numbers.

$(1)$	$a^{x} \times a^{y}$	$(a)$	$a^{x - y}$
$(2)$	$a^{x} \div a^{y}$	$(b)$	$a^{xy}$
$(3)$	$(a^{x})^{y}$	$(c)$	$a^{x + y}$
$(4)$	$(ab)^{x}$	$(d)$	$\dfrac {a^{x}}{b^{x}}$
$(5)$	$\left (\dfrac {a}{b}\right )^{x}$	$(e)$	$a^{x}\times b^{x}$

0%
$1 - (c), 2 - (a) , 3 - (b), 4 - (e), 5 - (d)$
0%
$1 - (b), 2 - (c) , 3 - (a), 4 - (e), 5 - (d)$
0%
$1 - (a), 2 - (c) , 3 - (d), 4 - (e), 5 - (b)$
0%
$1 - (a), 2 - (b) , 3 - (c), 4 - (d), 5 - (e)$

Explanation

(1) If base is same then the product of two exponential numbers is same as base raised to the power equal to sum of exponents of each.

$\therefore a^{x}\times a^{y} = a^{x+y}$

(2) If base is same then the division of two exponential numbers is same as base raised to the power equal to difference of exponents of each.

$\therefore a^{x}\div a^{y} = a^{x-y}$

(3)

$(a^{x})^{y} = a^{xy}$

(4)

$(ab)^x = (a^{x}b^{x}) = a^{x}\times b^{x}$

(5)

$(\dfrac{a}{b})^{x} = a^{x}\div b^{x} = \dfrac{a^{x}}{b^{x}}$

$\left(\dfrac{3}{2}\right)^2 \times \left(\dfrac{3}{2}\right)^{a+5}=\left(\dfrac{3}{2}\right)^8$ , then

$a =$ ______ .

0%
$-1$
0%
$0$
0%
$1$
0%
$2$

Explanation

$\left(\dfrac{3}{2}\right)^2\times\left(\dfrac{3}{2}\right)^{a+5}=\left(\dfrac{3}{2}\right)^8\Rightarrow \left(\dfrac{3}{2}\right)^{a+7}=\left(\dfrac{3}{2}\right)^8$
On comparing, we get,

$a+7=8 \Rightarrow a=1$

$\log {28}$ is same as_____

0%
$\log {16} +\log {12}$
0%
$\log {15} + \log {13}$
0%
$\log {2} + \log {14}$
0%
$\log {26} + \log {2}$

Explanation

$\log28$

$=\log(14\times2)$

$=\log14+\log2$ [Using:

$\log a+\log b =\log(a \times b)$ ]

Hence, C is the correct option.

$\log {10}$ is same as______

0%
$\log {6} + \log {3}$
0%
$\log {15} + \log {3}$
0%
$\log {21} - \log {3}$
0%
$\log {5} + \log {2}$

Explanation

$\log10$

$=\log(5\times2)$

$=\log5+\log2$ [Using:

$\log a+\log b =\log(a \times b)$ ]

Hence, D is the correct option.

Find the mantissa of

$\log 2.125$

0%
$1.3273$
0%
$2.3273$
0%
$0.3273$
0%
$32.2321$

Explanation

From Logarithmic table,

$\log2.125=0.3273$

Here, Characteristics

$=0,$ Mantissa

$=0.3273$

Hence, C is the correct option.

$\log{(2\times 3\times 5)}$ is equal to____

0%
$\log{2} + \log{3} + \log{5}$
0%
$\log{10}$
0%
$\log{30}$
0%
$\log{5}$

Explanation

$\log(2\times3\times5)=\log30$

$=\log2+\log3+\log5$

Hence, A and C are correct options.

Find the value :
i)

$\left(\dfrac{4}{5}\right)^3 \div \left(\dfrac{4}{5}\right)^2$
ii)

$4^7 \div 4^5$

0%
i) $\left(\dfrac{4}{5}\right)$
ii) $4$
0%
i) $\left(\dfrac{4}{5}\right)$
ii) $49$
0%
i) $\left(\dfrac{4}{5}\right)$
ii) $7$
0%
i) $\left(\dfrac{4}{5}\right)$
ii) $16$

Explanation

(1)

$\left ( \frac{4}{5} \right )^{3}\div \left ( \frac{4}{5} \right )^{2}$

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

$=\frac{4^{3}}{5^{3}}\times \frac{5^{2}}{4^{2}}$

$=\frac{4^{3}}{4^{2}}\times \frac{5^{2}}{5^{3}}$

Now apply law of exponents,

$\frac{x^{m}}{x^{n}}=x^{m-n}$

So, the expression is

$=4^{3-2}\times 5^{2-3}$

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

$=\frac{4}{5}$

(2)

$4^{7}\div 4^{5}$

$=4^{7}\times \frac{1}{4^{5}}$

$=\frac{4^{7}}{4^{5}}$

Now apply law of exponents,

$\frac{x^{m}}{x^{n}}=x^{m-n}$

So, the expression is

$=4^{7-5}$

$=4^{2}$

$=16$

Let

$\log_{\sqrt{8}}b=3\dfrac{1}{3},$ then find the value of

$b$ is

0%
$8$
0%
$16$
0%
$32$
0%
none of these

Explanation

Given,

$\log_{\sqrt{8}}b=3\dfrac{1}{3}$

$\implies b=(\sqrt{8})^{\large{\frac{10}{3}}}$

$\implies b=({2^3})^{\tfrac12\times\large{\frac{10}{3}}}$

$\implies b=2^5$

$\implies b=32$ .

$\log_ee^5$ is equal to-

0%
$2.5$
0%
$1.5$
0%
$2$
0%
$5$

Explanation

$\log_ee^5=5\log_ee=5*1=5$

$y = log \,x$ is a solution of

0%
$xy_2 = y_1$
0%
$xy_1 + y_2 = 0$
0%
$xy_2 + y_1 = 0$
0%
$xy_2 + y = 0$

Explanation

$y=log x$

$y_1=\dfrac{1}{x}$

$xy_1=1$

$xy_2+y_1=0$

$\therefore$

$y=log x$ is a solution of

$xy_2+y_1=0$ .

1104477_1208876_ans_150d2821117b422f809f1c31d61ba6c6.jpg

$a^{0}=1$ . Is it true or false?

0%
True
0%
False

Explanation

Any thing power zero is equal to

$1$

$\Rightarrow \therefore a^o=1$

Note: All values except

$a=0$ .

$\log\ (-2x)=2\log\ (x+1)$ , then

$x$ can be equal to

0%
$-2+\sqrt {3}$
0%
$-4+2\sqrt {3}$
0%
$-2-\sqrt {3}$
0%
$-4-2\sqrt {3}$

Explanation

We have,

$\log \left( -2x \right)=2\log \left( x+1 \right)$

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

Comparing both side and we get,

$-2x={{\left( x+1 \right)}^{2}}$

$\Rightarrow -2x={{x}^{2}}+1+2x$

$\Rightarrow {{x}^{2}}+4x+1=0$

Using quadratic formula and we get,

$x=\dfrac{-4\pm \sqrt{16-4\times 1\times 1}}{2\times 1}$

$x=\dfrac{-4\pm \sqrt{12}}{2}$

$x=\dfrac{-4\pm \sqrt{2\times 2\times 3}}{2}$

$x=\dfrac{-4\pm 2\sqrt{3}}{2}$

$x=-2\pm \sqrt{3}$

Hence, this is the answer.

The value of

$(0.125)^{\frac {2}{3}}$ is

0%
$2.5$
0%
$0.25$
0%
$0.025$
0%
$0.0025$

Explanation

$(0.125)^{\frac{2}{3}}$

$0.125$ could be written as

$(0.5)^3$

So given expression becomes'

$\Rightarrow (0.5^3)^{\frac{2}{3}}$

$= (0.5)^2$

$= 0.25$

State true or false:

${ \log }_{ a }{ x }^{ n }=n{ \log }_{ a }x$

0%
True
0%
False

Explanation

Step 1:

Let

$m = \log_a x$

Step 2: Write in exponent form

$x = a^m$

Step 3: Raise both sides to the power of

$n$

$x^n = ( a^m )^n$

Step 4: Convert back to a logarithmic equation

$\log_a x^n = mn$

Step 5: Substitute for

$m = \log_a x$

$\log_a x^n = n \log_a x$

$log_{8}m=3.5$ and

$log_{2}n=7$ , then the value of

$m$ in terms of

$n$ is

0%
$n\sqrt{n}$
0%
$2n$
0%
$n^2$
0%
$2n^2$

Explanation

Given,

$log_{8}m=3.5$ and

$log_{2}n=7$

Therefore,

$m=8^{3.5}$ and

$n=2^7$

$\implies m=8^{\dfrac{35}{10}}$

$\implies m=(2^7)^{3/2}$

$\implies m=n^{3/2}=n\sqrt{n}$

Which of the following real numbers is(are) non-positive?

0%
$log{ }_{ 0.3 }(\dfrac { \sqrt { 5 } +2 }{ \sqrt { 5 } -2 } )$
0%
$log{ }_{ 7 }(\sqrt { 83 } -9\quad )$
0%
$log{ }_{ 7\frac { \pi }{ 12 } }(cot\frac { \pi }{ 8 } \quad )$
0%
${ log }_{ 2 }\sqrt { 9.\sqrt [ 3 ]{ { 27 }^{ \frac { -5 }{ 3 } }.243{ }^{ \frac { -7 }{ 5 } } } }$

Explanation

From given,

we have,

$\log _{0.3}\left(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\right)$

$=\log _{0.3}\left(2+\sqrt{5}\right)-\log _{0.3}\left(\sqrt{5}-2\right)$

$=-2.39811$

${ \left( \dfrac { 2 }{ 3 } \right) }^{ 0 }=?$

0%
$\dfrac{3}{2}$
0%
$\dfrac{2}{3}$
0%
$1$
0%
$0$

Explanation

Given

${ \left( \dfrac { 2 }{ 3 } \right) }^{ 0 }$

By using tha law of exponents i.e.,

$\left(\dfrac{a}{b}\right)^0=1$

$\therefore { \left( \dfrac { 2 }{ 3 } \right) }^{ 0 }=1$

$5^{0}=5$

0%
True
0%
False

Explanation

False.
Using law of exponent

$=a^{0}=1$

$\Rightarrow 5^{0}=1$

$LHS\neq RHS$

State whether the following statement is true (T) or false (F):

$3^0 = (1000)^0$ .

0%
True
0%
False

Explanation

$3^0 =1$ and

$(1000)^0 =1$

$y$ is any non-zero integer, then

$y^0$ is equal to ....

0%
$1$
0%
$0$
0%
$-1$
0%
Not found

Explanation

Using the law of exponents,

$a^0 =1$

Therefore

$y^0=1$

$(-2)^{0}=2$

0%
True
0%
False

Explanation

${\textbf{Step -1: Write the equation}}{\textbf{.}}$

${\left( { - 2} \right)^0} = 2$

${\textbf{Step -2: According to law of indices}}{\textbf{.}}$

${\left( { - 2} \right)^0} = 1$

${\text{and }}$

${{ 1} } \neq 2$

${\textbf{Hence, the given equation is false}}{\textbf{.}}$

$(-6)^{0}=-1$

0%
True
0%
False

Explanation

False.
Using law of exponent

$=a^{0}=1$

$\Rightarrow (-6)^{0}=1$

$LHS\neq RHS$

$\log_{10} x = a$ and

$\log_{10} y = b$ , then

$10^{2b}$ in terms of

$y$ is equal to

0%
$y$
0%
$y^2$
0%
$y^3$
0%
$y^4$

Explanation

Given,

$\log _{10}x=a, \log _{10}y=b$

$\Rightarrow 10^b=y$

Consider,

$10^{2b}$

$=10^{b+b}$

$=10^b.10^b$

$=y.y$

$=y^2$ .

The value of the logarithmic function

$\log_2 \log_2 \log_2 16$ is equal to

0%
$0$
0%
$1$
0%
$2$
0%
$4$

Explanation

Using,

$\log_a x^y = y\log_a x$ and

$\log_a a = 1$

$\log _{ 2 }{ \log _{ 2 }{ \log _{ 2 }{ 16 } } } \\ =\log _{ 2 }{ \log _{ 2 }{ \log _{ 2 }{ { 2 }^{ 4 } } } } \\ =\log _{ 2 }{ \log _{ 2 }{ 4 } } \\ =\log _{ 2 }{ \log _{ 2 }{ { 2 }^{ 2 } } } \\ =\log _{ 2 }{ 2 } \\ =1\\$

The value of the expression

$\displaystyle\dfrac{(\log x - \log y)(\log x^{2} + \log y^{2})}{(\log x^{2} - \log y^{2})(\log x + \log y)}$ is equal to

0%
$0$
0%
$1$
0%
$\log\displaystyle \frac{x}{y}$
0%
$\log xy$

Explanation

Let

$E = \dfrac { (\log x-\log y)(\log { x }^{ 2 }+\log { y }^{ 2 }) }{ (\log { x }^{ 2 }-\log { y }^{ 2 })(\log x+\log y) }$

Using the formula,

$\log a^x = x\log a$

Therefore,

$E=\dfrac { (\log x-\log y)(2\log x+2\log y) }{ (2\log x-2\log y)(\log x+\log y) }$

$\Rightarrow E =\dfrac { (\log x-\log y)\times 2\times (\log x+\log y) }{ 2\times (\log x-\log y)\times (\log x+\log y) }$

$\Rightarrow E= 1$

Hence, option B is correct.

$\sqrt [4]{\sqrt [3]{2^{2}}}$ equal

0%
$2^{\dfrac {-1}{6}}$
0%
$2^{-6}$
0%
$2^{\dfrac {1}{6}}$
0%
$2^{6}$

Explanation

$\Rightarrow$

$\sqrt[4]{\sqrt[3]{2^2}}$

$\Rightarrow$

$[(2^2)^\tfrac{1}{3}]^{\tfrac{1}{4}}$

$\Rightarrow$

$(2)^{2\times \tfrac{1}{3}\times \tfrac{1}{4}}$

$\Rightarrow$

$2^{\tfrac{1}{6}}$

$\therefore$

$\sqrt[4]{\sqrt[3]{2^2}}=2^{\tfrac{1}{6}}$

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.