Tag: laws of exponents and powers

Quizzes (0) Questions (94)

Questions Related to laws of exponents and powers

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 } } $ 
$=1$.