Given exponential equation is

\(\displaystyle{6}^{{{1}-{4}{x}}}={5}^{{x}}\)

Taking natural log on both sides,

\(\displaystyle{\ln{{\left({6}^{{{1}-{4}{x}}}\right)}}}={\ln{{\left({5}^{{x}}\right)}}}\)

We know, \(\displaystyle{\ln{{\left({a}^{{b}}\right)}}}={b}{\ln{{\left({a}\right)}}}\)

\(\displaystyle{\left({1}-{4}{x}\right)}{\ln{{\left({6}\right)}}}={x}{\ln{{\left({5}\right)}}}\)

\(\displaystyle{\ln{{\left({6}\right)}}}-{4}{x}{\ln{{\left({6}\right)}}}={x}{\ln{{\left({5}\right)}}}\)

\(\displaystyle-{4}{x}{\ln{{\left({6}\right)}}}-{x}{\ln{{\left({5}\right)}}}=-{\ln{{\left({6}\right)}}}\)

\(\displaystyle-{x}{\left[{4}{\ln{{\left({6}\right)}}}+{\ln{{\left({5}\right)}}}\right]}=-{\ln{{\left({6}\right)}}}\)

\(\displaystyle{x}{\left[{4}{\ln{{\left({6}\right)}}}+{\ln{{\left({5}\right)}}}\right]}={\ln{{\left({6}\right)}}}\)

\(\displaystyle{x}={\frac{{{\ln{{\left({6}\right)}}}}}{{{\left[{4}{\ln{{\left({6}\right)}}}+{\ln{{\left({5}\right)}}}\right]}}}}\)

\(\displaystyle{x}\approx{0.2042}\)

Answer: The solution set is \(\{{\frac{{{\ln{{\left({6}\right)}}}}}{{{\left|{4}{\ln{{\left({6}\right)}}}+{\ln{{\left({5}\right)}}}\right|}}}}\}\) or \(\{0.2042\}\)