x = 2, y = 3 এবং z = 5

দেওয়া আছে, z = 5

প্রদত্ত রাশি $=lo{g}_5\left( \sqrt[3]{z}\sqrt{z}\right) = lo{g}_5\left( \sqrt[3]{5}\sqrt{5}\right)$

$= lo{g}_5\left( 5^{\frac{1}{3}}.5^{\frac{1}{2}}\right)$

$= lo{g}_5\left( 5^{\frac{1}{3}+\frac{1}{2}}\right)$

$= lo{g}_5\left( 5^{\frac{5}{6}}\right)$

$= \frac{5}{6} lo{g}_5 5$

$= \frac{5}{6} \times 1$

$= \frac{5}{6}$

নির্ণেয় মান: $\frac{5}{6}$

দেওয়া আছে, y = 3

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

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

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

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

$=3^{2a-a^2+1+a^2-1-2a-2+2}=3^0=1$

নির্ণেয় সরলমান 1.

দেওয়া আছে,

x = 2 y = 3 এবং z=5

বামপক্ষ = $\left({\log }_{10}\sqrt{y^3}+{\log }_{10 }{x}^3-{\log }_{10 }\sqrt{x^3 z^3}\right) ÷ lo{g}_{10} 1.2$

= $\frac{\left({\log }_{10}\sqrt{3^3}+{\log }_{10 }{2}^3-{\log }_{10 }\sqrt{2^3 5^3}\right)}{lo{g}_{10} 1.2 }$

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

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

= $\frac{\frac{3}{2}{\log }_{10}\left\{\left(3\times 4\right)÷10\right\}}{{\log }_{10} 1.2 }$

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

= $\frac{3}{2}$ ডানপক্ষ

$\left({\log }_{10}\sqrt{y^3}+{\log }_{10 }{x}^3-{\log }_{10 }\sqrt{x^3 z^3}\right) ÷ lo{g}_{10} 1.2$ = $\frac{3}{2}$ (দেখানো হলো)