Question

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

Answer 1

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

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

or $\frac{logx}{log2}+\frac{logx}{{log2}^2}+\frac{logx}{{log2}^4}=\frac{21}{4}$

or $logx(\frac{1}{log2}+\frac{1}{2logx}+\frac{1}{4log2})=\frac{21}{4}$

or $\frac{logx}{log2}(1+\frac{1}{2}+\frac{1}{4})=\frac{21}{4}$

or ${log}_{2}x\left(\frac{4+2+1}{4}\right)=\frac{21}{4}$

or ${log}_{2}x\left(7\right)=2+3$

${log}_{2}x=3$

$x=2^3=8$