Question

If ${\log }_{2}x+{\log }_{4}x+{\log }_{16}x=\frac{21}{4}$, find $x$.

Answer 1

$lo{g}_{2}x+lo{g}_{4}x+lo{g}_{16}x=\frac{21}{4}$

$\Rightarrow \frac{logx}{log2}+\frac{logx}{log4}+\frac{logx}{log16}=\frac{21}{4}$

$\Rightarrow logx[\frac{1}{log2}+\frac{1}{2log2}+\frac{1}{4log2}]=\frac{21}{4}$

$\Rightarrow \frac{logx}{logx}[1+\frac{1}{2}+\frac{1}{4}]=\frac{21}{4}$

$\Rightarrow \left[\frac{logx}{log2}\right]\frac{(4+2+1)}{4}=\frac{21}{4}$

$\Rightarrow \left(logx\right)=\frac{21}{7}\times log2$

$\Rightarrow logx=3log2$

$\Rightarrow logx=log{2}^3$

$\Rightarrow x=2^3$

$\Rightarrow x=8$.