Tag: power and exponent

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Questions Related to power and exponent

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

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

  1. $8 x^{16}$
  2. $2 x^{12}$
  3. $8 x^{12}$
  4. $2 x^{16}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
${\left( 16{x}^{16} \right)}^{\cfrac{3}{4}} = {\left( {2}^{4} {x}^{16} \right)}^{\cfrac{3}{4}}$
$\Rightarrow \; = {\left( {2}^{4} \right)}^{\cfrac{3}{4}} {\left( {x}^{16} \right)}^{\cfrac{3}{4}}$

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

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

$\Rightarrow \; = 8{x}^{12}$
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

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

  1. $3$
  2. $9$
  3. $0.3$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Now,

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

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

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 $abc$?

  1. $-1$
  2. $1$
  3. $0$
  4. $3$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$x=[(x)^{\dfrac{1}{c}}]^{\dfrac{1}{ab}}\Rightarrow x^1=x^{\dfrac{1}{abc}}$
$\therefore\dfrac{1}{abc}=1\Rightarrow abc=1$

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

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

  1. 10001

  2. 10002

  3. 1001

  4. 1002
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. $5$
  2. $2$
  3. $7$
  4. $6$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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 

  1. $0$
  2. $1$
  3. $2^ {m}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } } $ 
$=1$.