Question

Solve the following equation:
${\log }_{16}x+{\log }_{4}x+{\log }_{2}x=7$

Answer 1

$\frac{\log x}{4\log 2}+\frac{\log x}{2\log 2}+\frac{\log x}{\log 2}=7\log x+2\log x+4\log x=28\log 27\log x=28\log 2\log x=4\log 2x=2^{4}x=16$