Question

Solve ${\log }_{16}x+{\log }_{4}x+{\log }_{2}x=7$

Answer 1

${\log }_{16}x+{\log }_{4}x+{\log }_{2}x=7$
$⟹\frac{1}{{\log }_{x}16}+\frac{1}{{\log }_{x}4}+\frac{1}{{\log }_{x}2}=7$ $[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$\frac{1}{{\log }_x{2}^4}+\frac{1}{{\log }_x{2}^2}+\frac{1}{{\log }_{x}2}=7$
$\frac{1}{4{\log }_{x}2}+\frac{1}{2{\log }_{x}2}+\frac{1}{{\log }_{x}2}=7$ $[\because n{\log }_{a}M={\log }_a{M}^n]$
$[\frac{1}{4}+\frac{1}{2}+1]\frac{1}{{\log }_{x}2}=7⟹\left[\frac{7}{4}\right]\frac{1}{{\log }_{x}2}=7$
$\frac{1}{{\log }_{x}2}=7\times \frac{4}{7}$
${\log }_{2}x=4$ $[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
$2^4=x$ (exponential form)
$\therefore x=16$