Tag: power and exponent

Questions Related to power and exponent

Simplify: $( 16x ^{16} )^{\dfrac{3}{4}}$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $8 x^{16}$

  2. $2 x^{12}$

  3. $8 x^{12}$

  4. $2 x^{16}$

Show answer
Correct Option: C
Explanation:
${\left( 16{x}^{16} \right)}^{\cfrac{3}{4}} = {\left( {2}^{4} {x}^{16} \right)}^{\cfrac{3}{4}}$
$\Rightarrow \; = {\left( {2}^{4} \right)}^{\cfrac{3}{4}} {\left( {x}^{16} \right)}^{\cfrac{3}{4}}$

$\Rightarrow \; = {2}^\left( {4 \times \cfrac{3}{4}} \right)  {x}^\left({16 \times \cfrac{3}{4}} \right)$

$\Rightarrow \; = {2}^{3} \times {x}^{12}$

$\Rightarrow \; = 8{x}^{12}$

The value of $(0.243)^{0.2}\times (10)^{0.6}$ is 

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $3$

  2. $9$

  3. $0.3$

  4. None of these

Show answer
Correct Option: A
Explanation:

Now,

$(0.243)^{0.2}\times (10)^{0.6}$
$=\left(\dfrac{243}{1000}\right)^{\dfrac{1}{5}}\times (10)^{\dfrac{3}{5}}$
$=\left(\dfrac{3^5}{10^3}\right)^{\dfrac{1}{5}}\times (10)^{\dfrac{3}{5}}$
$=3\times (10)^{-\dfrac{3}{5}+\dfrac{3}{5}}$
$=3$.

Find the value of ${(6561)^{0.25}}$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $3$

  2. $7$

  3. $9$

  4. $19$

Show answer
Correct Option: C
Explanation:

$ \Rightarrow 6561 = {3^8}$

${\left( {6561} \right)^{0.25}} = {\left( {6561} \right)^{1/4}}$

$ = {\left( {{3^8}} \right)^{1/4}}$

$ = {\left( 3 \right)^{8/4}}$

$ = {3^2} = 9$

Solve:

$x^{\sqrt{x}}=\sqrt{x^x}$.

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $2$

  2. $3$

  3. $4$

  4. none of these

Show answer
Correct Option: C
Explanation:

Given,

$x^{\sqrt{x}}=\sqrt{x^x}$
or, $x^{\sqrt{x}}={x^\dfrac{x}{2}}$
or, $\sqrt{x}=\dfrac{x}{2}$
Squaring both sides we get,
$x=\dfrac{x^2}{4}$
Since $x\ne 0$ so $x=4$.

The difference between $3^{3^3}$ and $(3^3)^3$

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $3^{27}-3^9$

  2. $0$

  3. $27^3-3^{27}$

  4. $3^{18}(3^9-1)$

Show answer
Correct Option: A
Explanation:

We have,

$3^{3^3}=3^{(3^3)}=3^{27}$ and $(3^3)^3=(3^{3\times 3})=3^9$.
So the difference between them is $3^{27}-3^9$.

${\left( { - 2} \right)^{ - 5}}{\left( { - 2} \right)^6}$ is equals to

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $2$

  2. $-2$

  3. $-5$

  4. $6$

Show answer
Correct Option: B
Explanation:

Given that:

$(-2)^{-5}(-2)^{6}$
$=(\dfrac{-1}{2})^{5}(-2)^{6}$
$=(\dfrac{-1}{32})({64})$
$=(-1)(2)$
$=(-2)$

Hence, the value $-2$.

If $x = {y^{\frac{1}{a}}},\,y = {z^{\frac{1}{b}}}\,\,{\text{and}}\,\,z = {x^{\frac{1}{c}}}\,{\text{where}}\,x \ne 1,y \ne 1,\,z \ne 1$, then what is the value of $abc$?

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $-1$

  2. $1$

  3. $0$

  4. $3$

Show answer
Correct Option: B
Explanation:

Given $x=y^{\dfrac{1}{a}}.....(1)$

$y=z^{\dfrac{1}{b}}.....(2)$
$z=x^{\dfrac{1}{c}}.....(3)$
Putting the value of y from equation (2) in equation (1)
$x=[(z)^{\dfrac{1}{b}}]^{\dfrac{1}{a}}\Rightarrow x=z^{\dfrac{1}{ab}}$
Putting the value of z from equation (3) in the above equation
$x=[(x)^{\dfrac{1}{c}}]^{\dfrac{1}{ab}}\Rightarrow x^1=x^{\dfrac{1}{abc}}$
$\therefore\dfrac{1}{abc}=1\Rightarrow abc=1$

The value of $\frac{{{{100}^{98}} + {{100}^{100}}}}{{{{100}^{98}}}} + 1$ is equal to_____

maths power and exponent power of powers laws of exponents and powers law of indices

  1. 10001

  2. 10002

  3. 1001

  4. 1002

Show answer
Correct Option: B
Explanation:

The value of

  $ \dfrac{{{100}^{98}}+{{100}^{100}}}{{{100}^{98}}}+1 $

 $ =\dfrac{{{100}^{98}}}{{{100}^{98}}}+\dfrac{{{100}^{100}}}{{{100}^{98}}}+1 $

 $ =1+{{100}^{100-98}}+1 $

 $ ={{100}^{2}}+2 $

 $ =10000+2 $

 $ =10002 $


Hence, this is the answer. 

option (B) is correct.

Simplify the following $(3r^2)\times (9r^2)^{3/2} \div (27r^{-3})^{1/3}$ and find the power of $r$.

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $5$

  2. $2$

  3. $7$

  4. $6$

Show answer
Correct Option: D
Explanation:

We have,
$(3r^2)\times (9r^2)^{3/2}\div(27r^{-3})^{1/3}\$

$\Rightarrow (3r^2)\times ((3r)^2)^{3/2}\div(3^3r^{-3})^{1/3}\$
$\Rightarrow (3r^2)\times (3r)^{3}\div(3r^{-1})\$

$\Rightarrow (3r^2)\times (3^3r^3)\times(3^{-1}r)\\$
$\Rightarrow 27r^6$

So, the power of $r$ is $6$.

Hence, this is the answer.

The value of $\dfrac { { 2 }^{ m+3 }\times { 3 }^{ 2m-n }\times { 5 }^{ m+n+3 }\times { 6 }^{ n+1 } }{ { 6 }^{ m+1 }\times { 10 }^{ n+3 }\times { 15 }^{ m } } $ is equal to 

maths power and exponent power of powers laws of exponents and powers law of indices

  1. $0$

  2. $1$

  3. $2^ {m}$

  4. $none\ of\ these$

Show answer
Correct Option: B
Explanation:

Now,

$\dfrac { { 2 }^{ m+3 }\times { 3 }^{ 2m-n }\times { 5 }^{ m+n+3 }\times { 6 }^{ n+1 } }{ { 6 }^{ m+1 }\times { 10 }^{ n+3 }\times { 15 }^{ m } } $ 
$=\dfrac { { 2 }^{ m+3 }\times { 3 }^{ 2m-n }\times { 5 }^{ m+n+3 }\times(2^{n+1}\times { 3 }^{ n+1 }) }{ (2^{m+1}\times { 3 }^{ m+1 })\times (2^{n+3}\times { 5 }^{ n+3 })\times (3^{m}\times { 5 }^{ m }) } $ 
$=\dfrac { { 2 }^{ m+n+4 }\times { 3 }^{ 2m+1 }\times { 5 }^{ m+n+3 } }{ { 2 }^{ m+n+4 }\times { 3 }^{ 2m+1 }\times { 5 }^{ mm+n+3 } } $ 
$=1$.