Question

Solve for $x:{log}_2+{log}_{4}x+{log}_{16}x=\frac{21}{4}$.

Answer 1

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

$\Rightarrow lo{g}_{2}x+\frac{1}{2}lo{g}_{2}x+\frac{1}{4}lo{g}_{2}x=\frac{21}{4}$

$\Rightarrow \frac{1}{4}(4lo{g}_{2}x+2lo{g}_{2}x+lo{g}_{2}x)=\frac{21}{4}$

$\Rightarrow 7lo{g}_{2}x=21$

$\Rightarrow lo{g}_{2}x=3$

$\Rightarrow x=2^3=8$