Tag: laws of exponents and powers

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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

Simplicity
$\left[ \left{ \left( 625 \right) ^{ -\dfrac { 1 }{ 2 } } \right} ^{ -\dfrac { 1 }{ 4 } } \right] $

  1. $\dfrac{1}{\sqrt5}$
  2. $\sqrt5$

  3. 5

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

$\begin{array}{l}\left[ {{{\left( {{{\left( {625} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right] = \left[ {{{\left( {{{\left( {{{25}^2}} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = \left[ {{{\left( {{{25}^{ - 1}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = {25^{\frac{1}{4}}}\ = {5^{2 \times \frac{1}{4}}}\ = {5^{\frac{1}{2}}}\ = \sqrt 5 \end{array}$

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

If $\displaystyle \log _{16} 8$ = $\displaystyle \frac {3}{m}$, then value of $m$ is equal to 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\log _{16}8=\dfrac{3}{m}$

$\therefore \dfrac{\log 8}{\log 16}=\dfrac{3}{m}$      ....($\log _ba=\dfrac{\log a}{\log b}$)

$\therefore \dfrac{\log 2^3}{\log 2^4}=\dfrac{3}{m}$   ...($\log a^b=a\log b$)

$\therefore \dfrac{3}{m}=\dfrac{3}{4}$ $ \Rightarrow m = 4$.
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Match the numbers in column-I with the rules in column- II

No Column-I No Column-II
1 30 a $n^3+n/2 $
2 63 b $3n^2+3$
3 66 c $n^3+4$
4 110 d $n^2-2n$
5 127 e $n^3-3n$
f $2n^2-1$


Which rule the number 30 follows?

  1. b

  2. c

  3. d

  4. e
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Substitute the value of $n=3$ in $b$
 $\Rightarrow  3n^2+3=3\times 3^2+3=27+3= 30 $
Hence, option 'A' is correct.

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

$\displaystyle (64)^{-\tfrac{1}{2}}-(-32)^{-\tfrac{4}{5}}=?$

  1. $\displaystyle \frac{1}{8}$
  2. $\displaystyle \frac{3}{8}$
  3. $\displaystyle \frac{1}{16}$
  4. $\displaystyle \frac{3}{16}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle (64) ^{-\tfrac{1}{2}}-(-32)^{-\tfrac{4}{5}}$
$=\left({2^6}\right)^{\frac{-1}{2}}-\left({(-2)^5}\right)^{\frac{-4}{5}}$
$=(2)^{ -\tfrac { 6 }{ 2 }  }-(-2)^{ -\tfrac { 4\times 5 }{ 5 }  }$
$=(2)^{ -3 }-(-2)^{ -4 }$
$\dfrac { 1 }{ 8 } -\dfrac { 1 }{ 16 } =\dfrac { 1 }{ 16 } $
Answer $C$ option, $ \cfrac { 1 }{ 16 } $

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

If $a^x=\sqrt{b},b^y = \sqrt [3]{c}$ and $c^z = \sqrt {a}$ then the value of $xyz$

  1. $\displaystyle \frac {1}{2}$
  2. $\displaystyle \frac {1}{3}$
  3. $\displaystyle \frac {1}{6}$
  4. $\displaystyle \frac {1}{12}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$a^x=\sqrt{b}=b^{1/2}$
$a=b^{1/2x}$
$b^y=\sqrt[3]{c}$
$b^y=c^{1/3}$
$b=c^{1/3y}$
$c^z=a^{1/2}$
$c=a^{1/2z}=b^{1/4xz}$
$c^1=c^{1/12xyz}$
$\displaystyle 1= \frac {1}{12xyz}$
$\displaystyle xyz = \frac {1}{12}$

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

Solve for x ; $\displaystyle \frac{2^{x-3}}{8^{-x}} = \frac{32}{4^{(1/2)x}}$

  1. $2\displaystyle \frac{1}{5}$
  2. $1\displaystyle \frac{1}{5}$
  3. $3\displaystyle \frac{1}{5}$
  4. $1\displaystyle \frac{3}{5}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

 $\displaystyle \frac{2^{x-3}}{8^{-x}} = \frac{32}{4^{(1/2)x}}$
$\frac { 2^{ x-3 } }{ 2^{ -3x } } =\frac { { 2 }^{ 5 } }{ 2^{ (2/2)x } } $\ 
${ 2 }^{ x-3+3x }={ 2 }^{ 5-x }$
$4x-3=5-x$
$5x=8$
$x=1\frac { 3 }{ 5 } $
Answer (D)  $1\frac { 3 }{ 5 } $

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

If $2^a\,>\,4^c\;and\;3^b\,>\,9^a\;and\;a,\,b,\,c$ all positive, then

  1. $c\,<\,a\,<\,b$
  2. $b\,<\,c\,<\,a$
  3. $c\,<\,b\,<\,a$
  4. $a\,<\,b\,<\,c$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$2^a\,>\,4^c\;\;\;\;\;\;3^b\,>\,9^a$
$2^a\,>\,2^{2c}\;\;\;\;\;3^b\,>\,3^{2a}$
$a\,>\,2c\;\;\;\;\;\;\;b\,>\,2a$
$\therefore\;a\,>\,c-(i)\;\;\therefore\;b\,>\,a-(ii)$
From (1) & (2), we have
$c\,<\,a\,<\,b$

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

Find m so that $\displaystyle \left ( \frac{11^{2}}{13^{2}} \right )^{-6}=\left ( \frac{13}{11} \right )^{m}$

  1. $-12$
  2. $-6$
  3. $6$
  4. $12$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \left ( \frac{11^{2}}{13^{2}} \right )^{-6}=\left ( \frac{11}{13} \right )^{-12}=\left ( \frac{13}{11} \right )^{12}$
Also,$\displaystyle \left ( \frac{13}{11} \right )^{12}=\left ( \frac{13}{11} \right )^{m}$
$\displaystyle \therefore m=12$