If $$\displaystyle \log x=n$$ then 2n is equal to

$$\displaystyle \log\left ( x^{2} \right )$$

$$\displaystyle \left ( \log x \right )^{2}$$

$$\displaystyle \log \left ( x+2 \right )$$

If $$\displaystyle x=\frac{y}{(1+a)^p}$$, then $$p$$ is equal to

$$\displaystyle\frac{\displaystyle\log_e{\left(\frac{y}{x}\right)}}{\log_e{(1+a)}}$$

$$\displaystyle\log{\left\{\frac{y}{x(1+a)}\right\}}$$

$$\displaystyle\log{\left\{\frac{y-x}{1+a}\right\}}$$

$$\displaystyle\frac{\log{y}}{\log{\{x(1+a)\}}}$$

MCQ Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 3

The value of

$(2^0 - 3^0) \times 4^2$ is---

0%
$16$
0%
$0$
0%
$-16$
0%
None of these

Explanation

$(2^0 - 3^0) \times 4^2\\ = (1 - 1) \times 4^2\\ = 0\times 16\\=0$

$\displaystyle \log x=n$ then 2n is equal to

0%
$\displaystyle \log\left ( x^{2} \right )$
0%
$\displaystyle \left ( \log x \right )^{2}$
0%
$\displaystyle \log \left ( x+2 \right )$
0%
$\displaystyle \log 2x$

Explanation

Given ,

$\log x = n$
We know that

$\log x^{2} = 2 \log x$

$\Rightarrow 2n = 2 \log x = \log x^{2}$
Or

$\log x^{2} = 2 \log x = 2n$ .

$a^0=1$ is true for all

$a$ except

0%
$-1$
0%
negative integers
0%
0
0%
1

Explanation

The correct answer is C

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

${ (-2) }^{ 0 }=1$

${ 0 }^{ 0 }$ is not defined.

${ 1 }^{ 0 }=1$

$\displaystyle \log _{x}y=100\: and\: \log _{2}x=10$ then the value of y is

0%
$\displaystyle 2^{1000}$
0%
$\displaystyle 2^{100}$
0%
$\displaystyle 2^{2000}$
0%
$\displaystyle 2^{10000}$

Explanation

$\because log_2 x=10$

$\therefore x=2^{10}$
Also,

$log _x y=100$

$\therefore y=x^{100}$
Now, put the value of x then

$y=2^{1000}$

$\displaystyle x=\frac{y}{(1+a)^p}$ , then

$p$ is equal to

0%
$\displaystyle\frac{\displaystyle\log_e{\left(\frac{y}{x}\right)}}{\log_e{(1+a)}}$
0%
$\displaystyle\log{\left\{\frac{y}{x(1+a)}\right\}}$
0%
$\displaystyle\log{\left\{\frac{y-x}{1+a}\right\}}$
0%
$\displaystyle\frac{\log{y}}{\log{\{x(1+a)\}}}$

Explanation

$\displaystyle\because x=\frac{y}{(1+a)^p}$

$\displaystyle\therefore(1+a)^p=\frac{y}{x}$

$\displaystyle\therefore p\log_e{(1+a)}=\log_e{\frac{y}{x}}$

$\displaystyle\therefore p=\frac{\displaystyle\log_e{\left(\frac{y}{x}\right)}}{\log_e{(1+a)}}$

Which of the following expresses zero law of exponents?

0%
$0^{x} = 0$
0%
$x^{0} = 1$
0%
$x^{1} = x$
0%
None of the above

Explanation

According to the zero law of exponents,

any non - zero number raised to the power of zero is equal to

$1$ .

$\therefore x^{0} = 1$ , where

$x\neq 0$

So, option

$B$ is correct.

$4\times 4^{10}$ is represented as:

0%
$4^{40}$
0%
$4^{10}$
0%
$4^{11}$
0%
$16^{10}$

Explanation

$4\times 4^{10} = 4^{1 + 10}$

$= 4^{11}$

So, option

$C$ is correct.

In simplified form,

$((3)^{0} + (5)^{0})^{0}$ is equal to:

0%
$2$
0%
$1$
0%
$0$
0%
$8$

Explanation

$(3^{0} + 5^{0})^{0} = (1 + 1)^{0}$

$= (2)^{0}$

$= 1$

So, option

$B$ is correct.

Find the value of

$(6)^{0} - (10)^{0}$

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

Explanation

$(6)^0 - (10)^0 = 1-1 = 0$

So, option

$D$ is correct.

The expression

$((2)^{0} + (3)^{0} + (5)^{0})^{0}$ is equal to _____

0%
$3$
0%
$1$
0%
$10$
0%
$0$

Explanation

$((2)^{0} + (3)^{0} + (5)^{0})^{0} = (1 + 1 + 1)^{0}$

$= 3^{0}$

$= 1$

So, option

$B$ is the correct answer.

The value of

$21^{0}$ is _____.

0%
$0$
0%
$21$
0%
$1$
0%
$-21$

Explanation

$21^{0} = 1$

$\because a^{0} = 1$

This is the law for zero exponent.

So, option

$C$ is correct.

Which of the following expresses product law of exponents?

0%
$(a^{m})^{n} = a^{mn}$
0%
$a^{m}\times a^{n} = a^{m+n}$
0%
$a^{m} \div a^{n} = a^{m-n}$
0%
$a^{m}\div b^{m} = \left (\dfrac {a}{b}\right )^{m}$

Explanation

The product law of exponents states that,

when multiplying two powers that have the same base,

we can add their exponents.

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

So, option

$B$ is correct.

The value of

$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$ is _____

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

Explanation

$((6)^{0} + (16)^{0}) \div ((7)^{0} + (17)^{0})$

$= (1 + 1)\div (1 + 1)$

$= 2\div 2$

$= 1$

So, option

$C$ is correct.

The value of

$(100)^{0}$ is _____.

0%
$1$
0%
$0$
0%
$100$
0%
$1000$

Explanation

According to the law of power of

$0$ ,

$(100)^{0} = 1$ .

So, option

$A$ is correct.

Evaluate:

$({(10)^0} + (12)^{0})\times (18)^{0}$

0%
$1$
0%
$0$
0%
$2$
0%
$18$

Explanation

$((10)^{0} + (12)^{0})\times (18)^{0} = (1 + 1) \times 1$

$= (2)\times 1$

$= 2$

So. option

$C$ is correct.

Given that

$2^h\times 2^3 = 2^9$ , find the value of h.

0%
$3$
0%
$6$
0%
$8$
0%
$12$

Explanation

$2^{h}\times 2^{3}=2^{9}$

$\Rightarrow 2^{h+3}=2^{9}$

$\Rightarrow h+3=9$

$\Rightarrow \boxed{h=6}$

Find the correct expression, if

$\log _{ c }{ a } =x$ .

0%
${ x }^{ c }=a$
0%
${ a }^{ x }=c$
0%
${ c }^{ a }=x$
0%
${ c }^{ x }=a$
0%
None of these

Explanation

Given,

$\log _c a=x$

Taking antilog, we get

$c^x = a$ .

Which one of the following is the value of

$(101)^{0}$ ?

0%
$0$
0%
$101$
0%
$1010$
0%
$1$

Explanation

$101^{0} = 1$

So, option

$D$ is correct.

To simplify the following expression correctly, what must be done with the exponents?

${ { 5 }^{ a }\times { 5 }^{ b }\times }{ 5 }^{ c }$

0%
Add the exponents.
0%
Multiply the exponents.
0%
Divide the exponents
0%
Subtract the exponents.

Explanation

Since the base is same, the exponents can be added (Product rule of exponents).

$\therefore 5^a \times 5^b \times 5^c = 5^{(a+b+c)}$

Hence, option

$A$ is correct.

The value of

$(10)^{0} \times (20)^{0}\times (30)^{0}$ is:

0%
$1$
0%
$3$
0%
$60$
0%
$0$

Explanation

$(10)^{0} \times (20)^{0} \times (30)^{0} = 1\times 1\times 1 = 1$

So, option

$A$ is correct.

Which of the following is equal to

$1$ ?

0%
$(3)^{0} + (3)^{0}$
0%
$(3)^{0} - (3)^{0}$
0%
$(3)^{0}$
0%
$(3^{3})^{0} + (3^{0})^{3}$

Explanation

$\textbf{Step 1: Simplify the following expressions.}$

$(a)\ 3^0+3^0=1+1=2\neq 1$

$(b)\ 3^0-3^0=1-1=0\neq 1$

$(c)\ 3^0=1$

$(d)\ (3^3)^0+(3^0)^3=9^0+1^3=1+1=2\neq 1$

$\textbf{Hence, Option C is correct.}$

Simplify and give reasons:

${(-2)}^{7}$

0%
$-128$
0%
$128$
0%
$-28$
0%
None of these

Explanation

$(-2)^7$

$=(-1*2)^7$

$=(-1)^7(2)^7$

$\because (ab)^m=a^m*b^m$

$=-1*2^7$

$=-2^7$

$=-128$

${(1000)}^{9}\div {10}^{24}=$ ?

0%
$10000$
0%
$1000$
0%
$100$
0%
$10$
0%
None of the above

Explanation

Given Exp.=

$\cfrac { { \left( 1000 \right) }^{ 9 } }{ { 10 }^{ 24 } } =\cfrac { { \left( { 10 }^{ 3 } \right) }^{ 9 } }{ { 10 }^{ 24 } } =\cfrac { { 10 }^{ 27 } }{ { 10 }^{ 24 } } ={ 10 }^{ (27-24) }={ 10 }^{ 3 }=1000$

Simplify and give reasons:

${(-3)}^{-4}$

0%
$\cfrac{1}{81}$
0%
$\cfrac{-1}{81}$
0%
$\cfrac{3}{81}$
0%
None of these

Explanation

$(-3)^{-4}$

$=(-1*3)^{-4}$

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

$\because (ab)^m=a^m*b^m$

$=1*3^{-4}$

$=3^{-4}$

$\because a^{-m}=\dfrac1{a^m}$

$=(\dfrac13)^4$

$=\dfrac1{81}$

Simplify:

$\left( { 4 }^{ -1 }\times { 3 }^{ -1 } \right) \div { 6 }^{ -1 }$

0%
$\cfrac{1}{2}$
0%
$\cfrac{1}{4}$
0%
$\cfrac{2}{2}$
0%
None of these

Explanation

we know,

$a^m*b^m = (ab)^m$

so,

$4^{-1} * 3^{-1} = 12^{-1}$

we know,

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

so,

$\dfrac{12^{-1}}{6^{-1}}=2^{-1}=\dfrac{1}{2}$

Simplify the following:

${ (-2) }^{ 7 }\times { (-2) }^{ 3 }\times { (-2) }^{ 4 }$

0%
${(-2)}^{14}$
0%
${(-2)}^{24}$
0%
${(-2)}^{54}$
0%
None of these

Explanation

we know,

$a^{m}*a^{n}=a^{m+n}$

$\implies$

$a^{m}*a^{n}*a^{p}=a^{m+n+p}$

so,

${ (-2) }^{ 7 }\times { (-2) }^{ 3 }\times { (-2) }^{ 4 }$

$=(-2)^{7+3+4}$

$=(-2)^{14}$

Simplify and give reasons:

$\cfrac { { 3 }^{ -2 } }{ 3 } \times \left( { 3 }^{ 0 }-{ 3 }^{ -1 } \right)$

0%
$\cfrac{2}{81}$
0%
$\cfrac{1}{81}$
0%
$\cfrac{4}{81}$
0%
None of these

The logarithmic form of

${5}^{2}=25$ is

0%
$\log _{ 5 }{ 2 } =25$
0%
$\log _{ 2 }{ 5 } =25$
0%
$\log _{ 5 }{ 25 } =2$
0%
$\log _{ 25 }{ 5 } =2$

Explanation

$5^2=25$

Taking log with base

$5$ both sides, we get

$\log_55^2=\log_525$

$\Rightarrow \log_525=2\log_55$

$\Rightarrow \log_525=2$

$(\log_aa=1)$

Hence, C is the correct option.

Exponential form of

$\log_{4}8 = x$ is _____

0%
$x^{8} = 4$
0%
$x^{4} = 8$
0%
$4^{x} = 8$
0%
$8^{x} = 4$

Explanation

We have,

$\log_{4}{8}=x$

$\log_{b}{a}=x$ then

$a={b}^{x}$

So,

${4}^{x}=8$ is the correct answer.

Option

$C$ is the correct answer.

The exponential form of

$\log _{ 2 }{ 16 } =4$ is

0%
${2}^{4}=16$
0%
${4}^{2}=16$
0%
${2}^{16}=4$
0%
${4}^{16}=2$

Explanation

Since,

$\log_2 16=4$

$\implies \log$ of

$16$ to the base

$2$ is

$4$

$\Rightarrow 16=2^4$

Hence, A is the correct option.

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Logarithm And Antilogarithm Quiz 3 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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