Find the values of x in each of the following:

Solution:

From the following we have to find the value of x

(i) Given $2^{5 x} \div 2^{x}=\sqrt[5]{2^{20}}$

By using rational exponents $\frac{a^{m}}{a^{n}}=a^{m-n}$

$2^{5 x-x}=2^{20 \times \frac{1}{5}}$

On equating the exponents we get,

$5 x-x=4$

$4 x=4$

$x=\frac{4}{4}$

$x=1$

The value of $x$ is 1

(ii) Given $\left(2^{3}\right)^{4}=\left(2^{2}\right)^{x}$

$\begin{aligned} 2^{3 \times 4} &=2^{2 \times x} \\ 2^{12} &=2^{2 x} \end{aligned}$

On equating the exponents 

$\begin{aligned} 12 &=2 x \\ \frac{12}{2} &=x \\ 6 &=x \end{aligned}$

Hence the value of $x$ is 6

(iii) Given $\left(\frac{3}{5}\right)^{x}\left(\frac{5}{3}\right)^{2 x}=\frac{125}{27}$

Comparing exponents we have,

$2 x-x=3$

$1 x=3$

$x=\frac{3}{1}$

$x=3$

Hence the value of x is 

(iv) Given $5^{x-2} \times 3^{2 x-3}=135$

$5^{x-2} \times 3^{2 x-3}=5 \times 3^{3}$

On equating the exponents of 5 and 3 we get,

$x-2=1$

$x=3$

And,

$2 x-3=3$

$2 x=6$

$x=\frac{6}{2}$

$x=3$

The value of x is 

(v) Given $2^{x-7} \times 5^{x-4}=1250$

$2^{x-7} \times 5^{x-4}=2^{1} \times 625$

$2^{x-7} \times 5^{x-4}=2^{1} \times 5^{4}$

On equating the exponents we get 

$x-7=1$

$x=1+7$

$x=8$

And, 

$x-4=4$

$x=4+4$

$x=8$

Hence the value of x is 

(vi) $(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$

$\left(2^{2}\right)^{\frac{1}{3}\left(\frac{4 x+1}{2}\right)}=\left(\frac{1}{2}\right)^{5}$

$\Rightarrow 2^{\left(\frac{4 x+1}{3}\right)}=2^{-5}$

On comparing we get, 

$\frac{4 x+1}{3}=-5$

$\Rightarrow 4 x+1=-15$

$\Rightarrow 4 x=-16$

$\Rightarrow x=-4$

(vii) $5^{2 x+3}=1$

$5^{2 x+3}=5^{0}$

$\Rightarrow 2 x+3=0$

$\Rightarrow x=\frac{-3}{2}$

(viii) $(13)^{\sqrt{x}}=4^{4}-3^{4}-6$

$(13)^{\sqrt{x}}=\left(2^{2}\right)^{4}-3^{4}-6$

$\Rightarrow(13)^{\sqrt{x}}=2^{8}-3^{4}-6$

$\Rightarrow(13)^{\sqrt{x}}=256-81-6$

$\Rightarrow(13)^{\sqrt{x}}=169$

$\Rightarrow(13)^{\sqrt{x}}=(13)^{2}$

On comparing we get, 

$\sqrt{x}=2$

on squaring both sides we get,

$\mathrm{x}=4$

(ix) $\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\frac{125}{27}$

$\left(\sqrt{\frac{3}{5}}\right)^{x+1}=\left(\frac{5}{3}\right)^{3}$

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

On comparing we get, 

$\frac{x+1}{2}=-3$

$\Rightarrow x+1=-6$

$\Rightarrow x=-7$