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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?

$\frac{\log_2 e}{\log_2 n} = \frac{1}{\frac{\log_2 n}{\log_2 e}} = \frac{1}{\ln n}$

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    Yup, that's valid! – 2017-01-10
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    @fleablood good use of the 15 required characters – 2017-01-10
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    Yes. For any positive $A,B, x, y$ we have $\log_Ax/\log_Ay=\log_Bx/\log_By.$ From the def'n: $z=\log_Ax\iff A^z=x. $Always use the def'ns. They're not rules. They're tools. – 2017-01-10

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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.