Question

Solve : $2{\log }_{2}x+3{\log }_{4}x+5{\log }_{8}x=62$

Answer 1

Property: ${\log }_{a^k}b=\frac{1}{k}{\log }_{a}b$

Now,

$2{\log }_{2}x+3{\log }_{4}x+5{\log }_{8}x=62$

$2{\log }_{2}x+3{\log }_{(2{)}^2}x+5{\log }_{(2{)}^3}x=62$

$2{\log }_{2}x+\frac{3}{2}{\log }_{2}x+\frac{5}{3}{\log }_{2}x=62$

$\frac{31}{6}{\log }_{2}x=62$

${\log }_{2}x=\frac{62\times 6}{31}$

${\log }_{2}x=12$

$\Rightarrow x=2^{12}$