Question

Solve the following equation:

${\left(\frac{\log x}{2}\right)}^{{\log }^{2}x+\log x^2-2}=\log \sqrt{x}$

Answer 1

${\left(\frac{\log x}{2}\right)}^{{\log }^{2}x+{\log x}^2-2}=\log \sqrt{x}\Rightarrow {\left(\frac{\log x}{2}\right)}^{{\left(\log x\right)}^2+2\log x-2}=\left(\frac{\log x}{2}\right)\Rightarrow \frac{\log x}{2}=0or1or(\log x{)}^2+2\log x-2=1But\log xcannotbezero,since0=negativevalue=\infty Therefore,\log x=2or(\log x{)}^2+2\log x-3=0Therefore,x=100or(\log x-1)(\log x+3)=0\log x=1or\log x=-3Therefore,{\left(10\right)}^{-3},{10}^2,10$