<s>[INST] If $2^{2x-1}$ = $(\frac{1}{5})^x$ and $\log 2 = a$ prove that: $x=\frac{a}{a+1}$ This is my working out:
$2^{2x} \times 2^{-1} = (5^{-1})^x$
$\log 2^{2x-1} = \log 5^{-x}$
$2x-1 \log 2 = -x\log5$
$2x-1\times a = -x\log5$
At this point, I got stuck.
 [/INST] $$2^{2x-1}=5^{-x}$$
$$\begin{align}&\implies (2x-1)\log 2=-x \log 5 \\
&\implies\frac{1-2x}{x}=\frac {\log 5}{\log 2} \\
&\implies \frac 1x-2=\frac {\log 5}{\log 2} \\
&\implies \frac 1x=\frac{2\log 2+\log 5}{\log 2}\\
&\implies x=\frac{\log 2}{2\log 2+\log 5}=\frac{\log 2}{\log 20}\end{align}$$

If you mean $\lg$ instead of $\log$, then you have:
$$x=\frac{\lg 2}{\lg 20}=\frac{\lg 2}{1+\lg 2}.$$
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