Find $x$, if
(i) $ \left(\frac{1}{4}\right)^{-4} \times\left(\frac{1}{4}\right)^{-8}=\left(\frac{1}{4}\right)^{-4 x} $
(ii) $ \left(\frac{-1}{2}\right)^{-19}\div\left(\frac{-1}{2}\right)^{8}=\left(\frac{-1}{2}\right)^{-2 x+1} $
(iii) $ \left(\frac{3}{2}\right)^{-3} \times\left(\frac{3}{2}\right)^{5}=\left(\frac{3}{2}\right)^{2 x+1} $
(iv) $ \left(\frac{2}{5}\right)^{-3} \times\left(\frac{2}{5}\right)^{15}=\left(\frac{2}{5}\right)^{2+3 x} $
(v) $ \left(\frac{5}{4}\right)^{-x}\div\left(\frac{5}{4}\right)^{-4}=\left(\frac{5}{4}\right)^{5} $
(vi) $ \left(\frac{8}{3}\right)^{2 x+1} \times\left(\frac{8}{3}\right)^{5}=\left(\frac{8}{3}\right)^{x+2} $

To do:  

We have to find the value of $x$ in each case.

Solution:

We know that,

$a^m \times a^n=a^{m+n}$

$a^m \div a^n=a^{m-n}$

Therefore,

(i) \( \left(\frac{1}{4}\right)^{-4} \times\left(\frac{1}{4}\right)^{-8}=\left(\frac{1}{4}\right)^{-4 x} \)

$(\frac{1}{4})^{-4-8}=(\frac{1}{4})^{-4 x}$

$(\frac{1}{4})^{-12}=(\frac{1}{4})^{-4 x}$

Comparing both sides, we get,

$-4 x=-12$

$\Rightarrow x=\frac{-12}{-4}$

$x=3$

The value of $x$ is $3$.

(ii) \( \left(\frac{-1}{2}\right)^{-19}\div\left(\frac{-1}{2}\right)^{8}=\left(\frac{-1}{2}\right)^{-2 x+1} \)

$(\frac{-1}{2})^{-19-8}=(\frac{-1}{2})^{-2 x+1}$

$(\frac{-1}{2})^{-27}=(\frac{-1}{2})^{-2 x+1}$

Comparing both sides, we get,

$-2 x+1=-27$

$-2 x=-27-1$

$-2x=-28$

$x=\frac{-28}{-2}$

$x=14$

The value of $x$ is $14$.

(iii) \( \left(\frac{3}{2}\right)^{-3} \times\left(\frac{3}{2}\right)^{5}=\left(\frac{3}{2}\right)^{2 x+1} \)

$(\frac{3}{2})^{-3+5}=(\frac{3}{2})^{2 x+1}$

$(\frac{3}{2})^{2}=(\frac{3}{2})^{2 x+1}$

Comparing both sides, we get,

$2 x+1=2$

$2 x=2-1$

$2x=1$

$x=\frac{1}{2}$

The value of $x$ is $\frac{1}{2}$. 

(iv) \( \left(\frac{2}{5}\right)^{-3} \times\left(\frac{2}{5}\right)^{15}=\left(\frac{2}{5}\right)^{2+3 x} \)

$(\frac{2}{5})^{-3+15}=(\frac{2}{5})^{2+3 x}$

$(\frac{2}{5})^{12}=(\frac{2}{5})^{2+3 x}$

Comparing both sides, we get,

$2+3 x=12$

$3 x=12-2$

$3x=10$

$x=\frac{10}{3}$

The value of $x$ is $\frac{10}{3}$. 

(v) \( \left(\frac{5}{4}\right)^{-x}\div\left(\frac{5}{4}\right)^{-4}=\left(\frac{5}{4}\right)^{5} \)

$(\frac{5}{4})^{-x-(-4)}=(\frac{5}{4})^{5}$

Comparing both sides, we get,

$-x+4=5$

$-x=5-4$

$x=-1$

The value of $x$ is $-1$.  

(vi) \( \left(\frac{8}{3}\right)^{2 x+1} \times\left(\frac{8}{3}\right)^{5}=\left(\frac{8}{3}\right)^{x+2} \)

$(\frac{8}{3})^{2 x+1+5}=(\frac{8}{3})^{x+2}$

Comparing both sides, we get,

$2 x+6=x+2$

$2 x-x=2-6$

$x=-4$

The value of $x$ is $-4$.  

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Updated on: 10-Oct-2022

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