x = 2, y = 3 এবং z = 5

প্রদত্ত রাশি$={\log }_5 \left(\sqrt[3]{z}. \sqrt{z}\right)$

$={\log }_7 \left(5^{\frac{1}{3} }{7}^{\frac{1}{2} }\right) = {\log }_7\left(5^{\frac{1}{3}+\frac{1}{2}}\right)$

$=lo{g}_5{5}^{\frac{2+3}{6}}=lo{g}_5{5}^{\frac{5}{6}}=\frac{5}{6}lo{g}_{5}5=\frac{5}{6}\times 1=\frac{5}{6}$

নির্ণেয় মান $\frac{5}{6}$

দেওয়া আছে, y = 3

প্রদত্ত রাশি = $\frac{y^{a+1}}{{\left(y^a\right)}^{a-1}}÷\frac{{\left(3y\right)}^{a+1}}{{\left(y^{a+1}\right)}^{a-1}}\times \frac{1}{y^{-2}}$

= $\frac{3^{a+1}}{{\left(3^a\right)}^{a-1}}÷\frac{{\left(3.3\right)}^{a+1}}{{\left(3^{a+1}\right)}^{a-1}}\times \frac{1}{3^{-2}}$

= $\frac{3^{a+1}}{3^{a^2-a}}÷\frac{9^{a+1}}{3^{a^2-1}}\times 3^2$

= $3^{a+1-a^2+a}÷\frac{{\left(3^2\right)}^{a+1}}{3^{a^2-1}}\times 3^2$

=$3^{a+1-a^2+a}÷\frac{3^{2a+2}}{3^{a^2-1}}\times 3^2$

= $3^{2a-a^2+1}\times \frac{3^{a^2-1}}{3^{2a+2}}\times 3^2$

= $3^{2a-a^2+1}\times 3^{a^2-1-2a-2}\times 3^2$

= $3^{2a-a^2+1+a^2-1-2a-2+2}$

= $3^0$

= 1

নির্ণেয় সরলমান 1

দেওয়া আছে, x = 2, y = 3 এবং z = 5

বামপক্ষ  $\left(\log \sqrt{y^3} + \log x^3-\log \sqrt{x^3{z}^3} \right)+ \log 1.2$

$=\frac{\log \sqrt{3^3}+\log 2^3-\log \sqrt{2^3\times 5^3}}{log 1.2}$

$=\frac{\log 3^{{\displaystyle \frac{3}{2}}}+\log {\left(\sqrt{4}\right)}^3-\log \sqrt{{10}^3}}{log 1.2}$

$=\frac{\frac{3}{2} \log 3+\log 4^{{\displaystyle \frac{3}{2}}}-\log {10}^{{\displaystyle \frac{3}{2}}}}{\log 1.2}$

$=\frac{\frac{3}{2} \log 3+\frac{3}{2} \log 4-\frac{3}{2} \log 10}{\log 1.2}$

$=\frac{\frac{3}{2} \log \left(12÷10\right)}{\log 1.2}$

= $\frac{3}{2}$

ডানপক্ষ

$\therefore$ $\left(\log \sqrt{y^3} + \log x^3-\log \sqrt{x^3{z}^3} \right)+ \log 1.2 = \frac{3}{2}$ (দেখানো হলো)