Question

Solve for $x$:

$4^{{\log }_{9}3}+9^{{\log }_{2}4}={10}^{{\log }_{x}83}$.

Answer 1

Given,

$4^{{\log }_{9}3}+9^{{\log }_{2}4}={10}^{{\log }_{x}83}$

or, $4^{{\log }_{3^2}3}+9^{{\log }_2{2}^2}={10}^{{\log }_{x}83}$

or, $4^{\frac{1}{2}\times {\log }_{3}3}+9^{2\times {\log }_{2}2}={10}^{{\log }_{x}83}$

or, $4^{\frac{1}{2}}+9^2={10}^{{\log }_{x}83}$ [ Since ${\log }_{a}a=1$ for $a>0$]

or, $83={10}^{{\log }_{x}83}$

Taking logarithm with respect to the base $10$,

${\log }_{10}83={\log }_{x}83$

or, $x=10$.