Question

Solve the following equation:
$(\sqrt{x}{)}^{{\log }_{2}x-1}=3$

Answer 1

$(\sqrt{x}{)}^{{log}_{2}x-1}=3\Rightarrow \frac{{log}_{2}x-1}{2}={log}_{x}3$

Let $logx=a$

$\frac{a}{log2}-1=\frac{2log3}{a}\Rightarrow a^2-alog2-2log3log2=0\Rightarrow a=\frac{log2+\sqrt{{\left(log2\right)}^2+8log3log2}}{2}\Rightarrow logx=\frac{log2+\sqrt{{\left(log2\right)}^2+8log3log2}}{2}\Rightarrow x={10}^{\left(\frac{log2+\sqrt{log2log(2\times 3^8)}}{2}\right)}.$