In this item, I will show how the basic logarithmic rules, including the Change of Base formula, follow from this equivalency:

\[\log_b m = n \Leftrightarrow b^n = m\]

For the ease of reading, I’ll generally use the natural base (\(e\)) and the natural logarithm (\(\ln\)). However, everything here applies to all valid bases (\(b > 0, b \ne 1\)).

Also, note these are not fully rigorous, complete proofs.

To demonstrate: \(e^{\ln x} = x\).

First, note \(\log_b x = a \Leftrightarrow b^a = x\).

So \(\color{blue} b^{\color{red} {\log_b x}} = \color{green} x \Leftrightarrow \log_{\color{blue} b} \color{green} x = \color{red} {\log_b x}\). Since the latter is always true, the former is also always true.

That’s the general case, so it’s also true for the specific case of the natural base.

To demonstrate: \(\ln(mn) = \ln m + \ln n\).

Consider \(mn = e^{\ln(mn)}\).

Now consider \(mn = m\cdot n = e^{\ln m}\cdot e^{\ln n}\).

Recall that \(b^p \cdot b^q = b^{p+q}\).

Hence \(e^{\ln m} \cdot e^{\ln n} = e^{\ln m + \ln n}\).

Thus \(e^{\ln(mn)} = e^{\ln m + \ln n}\). \(e^p = e^q \Leftrightarrow p = q\), so \(\ln(mn) = \ln m + \ln n\).

The demonstration that \(\ln(\frac{m}{n}) = \ln m – \ln n\) is nearly identical.

To demonstrate: \(\ln(m^n) = n\ln(m)\).

Recall that, for positive integer \(n\), \(m^n = m\cdot m \cdot \dotsm \cdot m (n \text{ times})\).

For instance, \(\ln(m^2) = \ln(m\cdot m) = \ln m + \ln m = 2 \ln m\).

This can be generalized to \(\ln(m^n) = n\ln(m)\).

We can further generalize to all real \(n\).

To demonstrate: \(\log_b m = \frac{\log m}{\log b}\).

First: \(\log_b m = a \Leftrightarrow b^a = m\).

Also: \(\frac{\log m}{\log b} = a \Rightarrow \log m = a \log b \Rightarrow \log m = \log (b^a)\).

Furthermore, \(\log m = \log (b^a) \Leftrightarrow m = b^a\).

Hence both \(\log_b m = a\) and \(\frac{\log m}{\log b} = a\) imply \(b^a = m\), and since \(a = a\), \(\log_b m = \frac{\log m}{\log b}\).