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On twitter, I saw a funny looking identity
\begin{equation}\label{eqn:logab:20}
\log_{ab} x = \frac{ \log_a x \log_b x}{\log_a x + \log_b x}.
\end{equation}
To verify this, let
\begin{equation}\label{eqn:logab:40}
\begin{aligned}
u &= \log_a x \\
v &= \log_b x.
\end{aligned}
\end{equation}
This means that
\begin{equation}\label{eqn:logab:60}
\log_{ab} x = \log_{ab} a^u = \log_{ab} b^v.
\end{equation}
We may rewrite either of these in terms of \( a b \), for example
\begin{equation}\label{eqn:logab:80}
\begin{aligned}
\log_{ab} x
&= \log_{ab} b^v \\
&= v \log_{ab} b \\
&= v \log_{ab} \frac{ab}{a} \\
&= v \lr{ 1 – \log_{ab} a },
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:logab:100}
u \log_{ab} a = v \lr{ 1 – \log_{ab} a },
\end{equation}
or
\begin{equation}\label{eqn:logab:120}
\lr{ u + v } \log_{ab} a = v,
\end{equation}
or
\begin{equation}\label{eqn:logab:140}
u \log_{ab} a = \frac{u v}{u + v},
\end{equation}
and since \( x = a^u \), our proof is complete.