Tag: laws of indices

Questions Related to laws of indices

If $\dfrac {p}{q} = \left (\dfrac {2}{3}\right )^{3} \div \left (\dfrac {3}{2}\right )^{-3}$, then the value of $\left (\dfrac {p}{q}\right )^{10}$ is _______.

maths indices negative indices law of indices laws of indices

  1. $1$

  2. $0$

  3. $-1$

  4. Cannot be determined

Show answer
Correct Option: A
Explanation:
Given, $\dfrac {p}{q} = \left (\dfrac {2}{3}\right )^{3} \div \left (\dfrac {3}{2}\right )^{-3}$
We see that, $\left (\dfrac{3}{2}\right )^{-3}$=$\left (\dfrac{2}{3}\right)^{3}$
So, $\left (\dfrac{2}{3}\right)^{3}$ / $\left (\dfrac{2}{3}\right)^{3}$ $= 1$ 
Thus, $\left(\dfrac {p}{q}\right)^{10}=1$
Hence, A is the right answer.

$\left { \left (\dfrac {3}{4}\right )^{-1} - \left (\dfrac {1}{4}\right )^{-1}\right }^{-1} = ?$

maths indices negative indices law of indices laws of indices

  1. $\dfrac {3}{8}$

  2. $\dfrac {-3}{8}$

  3. $\dfrac {8}{3}$

  4. $\dfrac {-8}{3}$

Show answer
Correct Option: B
Explanation:

We need to find value of $\left { \left (\dfrac {3}{4}\right )^{-1} - \left (\dfrac {1}{4}\right )^{-1}\right }^{-1} $
It can be written as $\left (\dfrac {4}{3} - 4\right)^{-1}$ $=$ $\left (\dfrac {-8}{3}\right)^{-1}$ $=$ $-\dfrac {3}{8}$

The value of $\left (\dfrac {32}{243}\right )^{-3/5}$ is _____.

maths indices negative indices law of indices laws of indices

  1. $\dfrac {27}{8}$

  2. $\dfrac {8}{27}$

  3. $\dfrac {16}{27}$

  4. $\dfrac {27}{16}$

Show answer
Correct Option: A
Explanation:
We need to find value of $\left (\dfrac {32}{243}\right )^{-3/5}$
It can be written as,
$\left(\dfrac {2^5}{3^5}\right)^{-3/5}$
$\Rightarrow \left (\dfrac {3}{2}\right)^{3} = \dfrac {27}{8}$

$\left {\left (\dfrac {1}{3}\right )^{-3} -\left (\dfrac {1}{2}\right )^{-3}\right } \div \left (\dfrac {1}{4}\right )^{-3} = ?$

maths indices negative indices law of indices laws of indices

  1. $\dfrac {19}{64}$

  2. $\dfrac {64}{19}$

  3. $\dfrac {27}{16}$

  4. $\dfrac {16}{27}$

Show answer
Correct Option: A
Explanation:
We need to find value of $\left \{\left (\dfrac {1}{3}\right )^{-3} -\left (\dfrac {1}{2}\right )^{-3}\right \} \div \left (\dfrac {1}{4}\right )^{-3} = ?$
$\left (\dfrac{1}{3}\right)^{-3}=3^{3}$
$\left (\dfrac{1}{2}\right)^{-3}=2^{3}$
$\left (\dfrac{1}{4}\right)^{-3}=4^{3}$
So, given expression becomes,
$\dfrac {3^{3}-2^{3}}{4^{3}}$ $=\dfrac {19}{64}$. 
Hence, A is the right option.

$\dfrac{1}{5}$ can also be expressed as:

maths indices negative indices law of indices laws of indices

  1. $5^{-1}$

  2. $\dfrac{1}{25}$

  3. $\dfrac{-1}{25}$

  4. None of these

Show answer
Correct Option: A
Explanation:

$\dfrac 15 = \dfrac {1}{5^1}=5^{-1}$

$-2^{-3}$ can also be expressed as:

maths indices negative indices law of indices laws of indices

  1. $\dfrac{1}{8}$

  2. $-\dfrac{1}{8}$

  3. $8$

  4. $-8$

Show answer
Correct Option: B
Explanation:

$-2^{-3} = \dfrac {1}{-2 \times -2 \times -2} = -\dfrac 18$


So, option B is correct.

The inverse of the function $f : R \rightarrow {x \in R : x < 1}$ given by $f(x)=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}},$ is 

maths indices negative indices law of indices laws of indices

  1. $\dfrac{1}{2} \log \dfrac{1+x}{1-x}$

  2. $\dfrac{1}{2} \log \dfrac{2+x}{2-x}$

  3. $\dfrac{1}{2} \log \dfrac{1-x}{1+x}$

  4. $none\ of\ these$

Show answer
Correct Option: A
Explanation:

Given $f(x)=\cfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}}$

So $y=\cfrac{e^x-e^{-x}}{e^{x}+e^{-x}}$
$\implies y=\cfrac{e^{2 x}-1}{e^{2 x}+1}$
$\implies y(e^{2 x}+1)=e^{2 x}-1$
$\implies e^{2 x}(y-1)=-y-1$
$\implies e^{2 x}=\cfrac{1+y}{1-y}$
$\implies 2 x=\log \cfrac{1+y}{1-y}$
$\implies x=\cfrac{1}{2}\log \cfrac{1+y}{1-y}$
$f^{-1}(x)=\cfrac{1}{2}\log\cfrac{1+x}{1-x}$

If $\sqrt [ 3 ]{ a+\sqrt { b }  } =7+4\sqrt { 3 } $, then $\sqrt [ 3 ]{ { a }^{ 2 }-b } =$

maths indices negative indices law of indices laws of indices

  1. $0$

  2. $1$

  3. $-1$

  4. $7$

Show answer
Correct Option: B
Explanation:

We have.

$ \sqrt[3]{a+\sqrt{b}}=7+4\sqrt{3} $

$ {{(a+\sqrt{b})}^{1/3}}=(7+4\sqrt{3}) $

$ a+\sqrt{b}={{(7+4\sqrt{3})}^{3}}\ \ ......\ \ (1) $


$ \text{Similarly,} $

$ a-\sqrt{b}={{(7-4\sqrt{3})}^{3}}\ \ ......\ \ (2) $


On multiplying (1) and (2) to. We get,

$ (a+\sqrt{b})(a-\sqrt{b})={{(7+4\sqrt{3})}^{3}}{{(7-4\sqrt{3})}^{3}} $

$ {{a}^{2}}-b={{\left[ (7+4\sqrt{3})(7-4\sqrt{3}) \right]}^{3}} $

$ {{a}^{2}}-b={{\left[ 49-16\times 3 \right]}^{3}} $

$ {{({{a}^{2}}-b)}^{1/3}}=(1) $

$ \sqrt[3]{{{a}^{2}}-b}=1 $


Hence, this is the answer

$(x+y)^{-1}(x^{-1}+y^{-1})$ is the reciprocal of

maths indices negative indices law of indices laws of indices

  1. $x+y$

  2. $xy$

  3. $\displaystyle\frac{1}{xy}$

  4. $\displaystyle\frac{x}{y}$

Show answer
Correct Option: B
Explanation:

$\displaystyle(x+y)^{-1}(x^{-1}+y^{-1})=\frac{1}{x+y}\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{x+y}\times\frac{x+y}{xy}=\frac{1}{xy}$

The value of ${1-[1-(1-n)^{-1}]^{-1}}^{-1}$ is

maths indices negative indices law of indices laws of indices

  1. $0$

  2. $1$

  3. $n$

  4. $\displaystyle\frac{1}{n}$

Show answer
Correct Option: C
Explanation:

$\displaystyle{1-[1-(1-n)^{-1}]^{-1}}^{-1}=\left[1-\left(1-\frac{1}{1-n}\right)^{-1}\right]^{-1}$

$\displaystyle=\left[1-\left(\frac{1-n-1}{1-n}\right)^{-1}\right]^{-1}=\left[1+\left(\frac{n}{1-n}\right)^{-1}\right]^{-1}=\left(1+\frac{1-n}{n}\right)^{-1}$

$\displaystyle=\left(\frac{n+1-n}{n}\right)^{-1}=\left(\frac{1}{n}\right)^{-1}=n$