Question

If ${\log }_{8}x+{\log }_{4}x+{\log }_{2}x=11$, find $x$.

Answer 1

${\log }_{8}x+{\log }_{4}x+{\log }_{2}x=11$

or, ${\log }_{2^3}x+{\log }_{2^2}x+{\log }_{2}x=11$

or, $\frac{1}{3}{\log }_{2}x+\frac{1}{2}{\log }_{2}x+{\log }_{2}x=11$

or, $\frac{11}{6}{\log }_{2}x=11$

or, ${\log }_{2}x=6$

or, $x=2^6$

or, $x=64$.