$P=\frac{5.2^x-2.2^{x-1}}{2^x-2^{x-1}}$, $Q=\frac{5^{m+2}-3.5^{m+1}}{5^{m+2}÷5}$ এবং $R=\frac{{\log }_{10} \sqrt{125}+{\log }_{10} 27-{\log }_{10}1000}{lo{g}_{10 }4.5}$

${\left(\sqrt{3}\right)}^{x+1}= {\left(\sqrt[3]{3}\right)}^{ 2x-1}$

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

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

বা, 4x - 2 = 3x + 3

বা, 4x - 3x = 3 + 2

$\therefore$ x = 5

নির্ণেয় মান 5.

$P=\frac{5.2^x-2.2^{x-1}}{2^x-2^{x-1}}$

$=\frac{5.2^x-2.2^x.2^1}{2^x-{\displaystyle \frac{2^X}{2}}}$

$=\frac{5.2^x-2^{1-1} .2^x}{2^x-{\displaystyle \frac{2^X}{2}}}$

$=\frac{2^x\left(5-2^0\right)}{2^x\left(1-{\displaystyle \frac{1}{2}}\right)}$

= $\frac{5-1}{{\displaystyle \frac{1}{2}}}$

= $4\times 2$

= 8

$Q=\frac{5^{m+2}-3.5^{m+1}}{5^{m+2}÷5}$

= $\frac{5^m.5^2-3.5^m.5}{{\displaystyle \frac{5^M.5^2}{5}}}$

= $\frac{5^M(5^2-15)}{5^M.5}$

= $\frac{25-15}{5}$

=$\frac{10}{5}=2$

বামপক্ষ = $P÷Q$$÷\sqrt{16}$

=  $P\times \frac{1}{Q}\times \frac{1}{\sqrt{16}}$

= $8\times \frac{1}{2}\times \frac{1}{4}$ = 1

$P÷Q÷\sqrt{16}=1$ (প্রমাণিত)

$R=\frac{{\log }_{10} \sqrt{125}+{\log }_{10} 27-{\log }_{10}1000}{lo{g}_{10 }4.5}$

$=\frac{{\log }_{10} {\left(5^3\right)}^{{\displaystyle \frac{1}{2}}}+{\log }_{10} 3^3-{\log }_{10}{\left({10}^3\right)}^{{\displaystyle \frac{1}{2}}}}{lo{g}_{10 }{\displaystyle \frac{45}{10}}}$

$=\frac{\frac{3}{2} {\log }_{10} 5+3 {\log }_{10} 3-\frac{3}{2} {\log }_{10}10}{lo{g}_{10 }\left(\frac{5\times 3^2}{10}\right)}$

$=\frac{\frac{3}{2}\left( {\log }_{10} 5+2 {\log }_{10} 3-1\right)}{ {\log }_{10} 5+2 {\log }_{10} 3- {\log }_{10}10}$

$=\frac{\frac{3}{2}\left( {\log }_{10} 5+2 {\log }_{10} 3-1\right)}{\left( {\log }_{10} 5+2 {\log }_{10} 3-1\right)}$

= $\frac{3}{2}$

বামপক্ষ $= R ÷ 9 = \frac{3}{2}÷ 9 = \frac{3}{2}\times \frac{1}{9}= \frac{1}{6}=$ডানপক্ষ

$\therefore$ $R ÷ 9 = \frac{1}{6}$ (দেখানো হলো)