Exponents and Logarithms

\(a\) to the power of 1 is therefore just simply \(a\).

\[a^0 = 1 \quad \text{for all } a \neq 0 \]

Depending on the context, \(0^0\) can be defined as 1 or undefined.

\[a^b \cdot a^c = a^{b+c} \] \[(ab)^c = a^b \cdot b^c \] \[(a^b)^c = a^{b \cdot c} \] \[\frac{a^b}{a^c} = a^{b-c} \] \[\left(\frac{a}{b}\right)^c = \frac{a^c}{b^c} \] \[a^{\frac{b}{c}} = \sqrt[c]{a^b} \] \[a^{-b} = \frac{1}{a^b} \]

Logarithms

The logarithm of \(b\) with base \(a\) is the power to which \(a\) must be raised to obtain \(b\), so answers the question of ”\(a\) to the power of what gives \(b\)?”

This is why the logarithm is only defined for a positive base \(a > 0\) and \(a \neq 1\), and a positive argument \(b > 0\).

\[\log_a{b} = c \iff a^c = b \] \[\log_a{a} = 1 \] \[\log_a{a^b} = b \]

Because of this, the logarithm of 1 with any base is 0.

\[\log_a{1} = 0 \] \[\log_a{b \cdot c} = \log_a{b} + \log_a{c} \] \[\log_a{\frac{b}{c}} = \log_a{b} - \log_a{c} \] \[\log_a{b^c} = c \cdot \log_a{b} \] \[- \log_a{b} = \log_a{\frac{1}{b}} \]

Change of base formula:

\[\log_a{b} = \frac{\log_c{b}}{\log_c{a}} \] \[\log_a{b} \cdot \log_c{a} = \log_c{b} \] \[\log_a{b} = \frac{1}{\log_b{a}} \] \[a^{\log_a{b}} = b \] \[\log_{a^b}{c} = \frac{1}{b} \cdot \log_a{c} \] \[\log_{a^b}{c^d} = \frac{d}{b} \cdot \log_a{c} \]