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.

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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.

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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.

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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 $n$.

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