Question

$\left(\frac{(81{)}^{\frac{1}{{\log }_{5}9}}+3^{\frac{3}{{\log }_{\sqrt{6}}3}}}{409}\right)\left(\right(\sqrt{7}{)}^{\frac{2}{{\log }_{25}7}}-\left(125{)}^{{\log }_{25}6}\right)=$

• A. $0$
• B. $1$
• C. $\sqrt{7}+2$
• D. $5+2\sqrt{7}$

Answer 1

The correct option is B $1$

We know that $a^{{\log }_{a}b}=b$

$\left(\frac{(9^2{)}^{{\text{log}}_{9}5}+3^{3{\text{log}}_3\sqrt{6}}}{409}\right)(7^{{\text{log}}_{7}25}-6^{\frac{3}{2}})=\left(\frac{25+6\sqrt{6}}{409}\right)(25-6\sqrt{6})$
$=\frac{625-216}{409}=1$