logarithm and exponent properties

Logarithm and exponent properties

2024-04-21

Exponent rules

The exponent rules appear when you visualize what’s being multiplied, then slap basic algebra on. With these two rules:

\begin{align} x^ax^b &= x^{a+b} \\ &= (x * x *... \text{a times}) \times (x * x *... \text{b times}) \notag \\ &= (x * \text{itself, } a+b \text{ times}) \notag \notag \\ (x^a)^b &= x^{ab} \\ &= (x* \text{itself, a times}) \times \text{itself, b times} \notag \\ &= (x * \text{itself, } a \times b \text{ times}) \notag \end{align}

We can quickly infer rooting, and then $\frac{power}{root}$ :

\begin{align} \sqrt[b]{x} &= x^{1/b} \quad \text{\ From rule 2, since} \\ (\sqrt[b]{x})^b &= (x^{1/b})^b = x^{(b/b)} \notag \\ \notag \\ x^{a/b} &= \sqrt[b]{x^a} \quad \text{\ From rule 3, since } \\ x^{a/b} &= (x^a)^{1/b} = \sqrt[b]{x^a} \notag \\ \end{align}

With rule 1, we can also infer that:

\begin{align} x^0 &= x^1 \div x^1 = 1 \quad \text{From rule 1} \\ \notag \\ x^{-a} &= x^0 \div x^a = \frac{1}{x^a} \quad \text{From rule 1 and 5} \\ \notag \\ a^xb^x &= (ab)^x \\ &= (a * a * ... \text{x times}) \times (b * b * ...\text{x times}) \notag \\ &= (ab * ab * ... \text{x times by associative property})\notag \end{align}

The logarithm is the exponent needed to get a base to a number, i.e.
$b^x=n \iff \log_b{n}=x$

By definition, $\log_b(b^x)=x$ since the exponent needed to get any $b$ to $b^{x}$ is… well, $x$ .

\begin{align} \log_b(mn) &= \log_bm + \log_bn \\ \text{Let } m = b^x \text{ and } n &= b^y \text{ so we can substitute:} \notag \\ \log_b(b^xb^y) &= \log_b(b^x) + \log_b(b^y) \notag \\ \log_b(b^{x + y}) &= \log_b(b^x) + \log_b(b^y) \notag \\ x + y &= x + y \quad \text{ by definition} \notag \end{align}

This is just the extension of above, but with the use of $n=b^y$ becoming $n^{-1}=b^{-y}$ .

\begin{align} \log_b(\frac{m}{n}) &= log_bm - log_bn \\ \log_b(mn^{-1}) &= log_bm - log_bn \notag \\ \log_b(b^x(b^y)^{-1}) &= \log_b(b^x) + log_b((b^y)^{-1}) \notag \\ \log_b(b^xb^{-y}) &= \log_b(b^x) + log_b(b^{-y}) \notag \\ \log_b(b^{x-y}) &= \log_b(b^x) + log_b(b^{-y}) \notag \\ x - y &= x + - y \notag \end{align}

\begin{align} \log_b(\sqrt[q]{m^p}) &= \frac{p}{q} \log_bm \\ \log_b(m^{\frac{p}{q}}) &= \frac{p}{q} \log_bm \notag \\ \log_b((b^x)^{\frac{p}{q}}) &= \frac{p}{q} \log_b(b^x) \notag \\ \log_b(b^{\frac{xp}{q}}) &= \frac{p}{q} \log_b(b^x) \notag \\ x \times \frac{p}{q} &= x \times \frac{p}{q} \notag \end{align}

Of course, we only use the principal solution of the root, since a logarithm cannot take a negative argument (its domain is $x>0$ ).

Any given logarithm can be written in another arbitrary-base logarithm.

\begin{align} \log_bm &= \frac{\log_nm}{\log_nb} \\ \log_b(b^x) &= \frac{\log_n(b^x)}{\log_nb} \notag \\ \log_b(b^x) &= \frac{x\log_nb}{log_nb} \quad \text{ By the power rule} \notag \\ x &= x \notag \end{align}