# The Logarithmic Rules

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 $$a = a$$, $$\log_b m = \frac{\log m}{\log b}$$.

Clio Corvid

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