$\begin{array}{l}
{5^x} = 1 \circ \,\,\, \Rightarrow \log {5^x} = \log 1 \circ \, \Rightarrow x\log 5 = 1\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x\log \frac{{1 \circ }}{2} = 1\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x(\log 1 \circ  - \log 2) = 1\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x(1 - \log 2) = 1\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow x = \frac{1}{{1 - \log 2}}
\end{array}$

$\begin{array}{l}
{2^{f(x)}} = 2 \circ \,\,\,\, \Rightarrow \log {2^{f(x)}} = \log \,\,2 \circ \,\,\, \Rightarrow f(x)\log 2 = \log (1 \circ  \times 2)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow f(x)\log 2 = \log 1 \circ  + \log 2\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow f(x)\log 2 = 1 + \log 2\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow f(x) = \frac{1}{{\log 2}} + 1
\end{array}$

$x = \frac{1}{{1 - \log 2}}\,\,\, \Rightarrow 1 - \log 2 = \frac{1}{x}\,\, \Rightarrow \log 2 = 1 - \frac{1}{x} = \frac{{x - 1}}{x}\,\,\, \Rightarrow \frac{1}{{\log 2}} = \frac{x}{{x - 1}}$

$f(x) = \frac{1}{{\log 2}} + 1\,\,\, \Rightarrow f(x) = \frac{x}{{x - 1}} + 1 = \frac{{x + x - 1}}{{x - 1}} = \frac{{2x - 1}}{{x - 1}}$