Question

Solve the following equations.
$x^{logx-3}={10}^{log\left(10/3\right)-1}.$

Answer 1

$x^{logx-3}={10}^{log\left(10/3\right)-1}$

Log on both sides

$(logx-3)logx=log\left(\frac{10}{3}\right)-1$

${\left(logx\right)}^2-3logx={log}^{10-log3-1}$

${\left(logx\right)}^2-3logx+log3=0$

$logx=\frac{3\pm \sqrt{9-4log3}}{2}$ [From Sreedhachrya Method]

$logx=\frac{3\pm 3.3}{2}$

$logx=\frac{6.3}{2}=3.15$ or $logx=\frac{-0.3}{2}$

$x={10}^{3.15}$ (Answer) $x={10}^{-0.15}$