We have to solve the following equation

\(\displaystyle{\left({\log{{\left({7}{x}\right)}}}\right)}{\log{{\left({x}\right)}}}={5}\)

By applying log rule \(\log c(ab)=\log c(a)+\log c(b)\) we get

\((\log(7x))\log(x)=5 \)

⟹ \((\log(x)+\log(7))\log(x)=5 \)

⟹ \((u+\log(7))u=5\),[where \(u=\log(x)\)]

⟹ \( u_2+\log(7)u−5=0\)

The above is a quadratic equation of the form \(au^2+bu+c=0\). By using quadratic formula we get

\(\displaystyle{u}=\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{2}}{a}=\frac{{-{\log{{7}}}\pm\sqrt{{{\log{{7}}}}}^{{2}}+{20}}}{{2}}\)

Since \(\displaystyle{u}={\log{{x}}}\), x=eu. Therefore, the required solutions are

\(\displaystyle{x}={e}^{{-{\log{{7}}}\pm\frac{\sqrt{{{\left({\log{{7}}}^{{2}}\right)}+{20}}}}{{2}}}}\)