Assuming we have an expression $\frac{\log_2 e}{\log_2 n}$ which we want to simplify.
My first way to do this would be applying the log rules:
$\log_n e$
But this is hardly simpler. Ideally we'd like to have a constant base, so:
$\log_2 e = \frac{1}{\ln 2}, \\
\log_2 n = \frac{\ln n}{\ln 2} \\
\frac{\log_2 e}{\log_2 n} = \frac{1}{\ln 2} \cdot \frac{\ln 2}{\ln n} = \frac{1}{\ln n}$
But that's a lot of calculations. Would the following be valid, too?
$\frac{\log_2 e}{\log_2 n} = \frac{1}{\frac{\log_2 n}{\log_2 e}} = \frac{1}{\ln n}$
Best Answer
Yes. It is valid.
Also note (assuming a, b, and c are all positive and "nice"):
$\log_a b =c \iff $
$a^c = b \iff $
$a = b^{\frac 1c}\iff $
$\frac 1c = \log_b a $
So $\log_b a = \frac 1 {\log_a b} $
...
It's all good.