# Step By Step Calculus » 7.3 - Logarithmic and Inverse Hyperbolic Functions

Synopsis
The exponential function a^xa^x is strictly decreasing if a\in(0,1)a\in(0,1) and strictly increasing if a>1a>1. The inverse of a^{x}a^{x} is called the logarithm with base aa, written \log_{a}(x)\log_{a}(x), which is also strictly decreasing if a\in(0,1)a\in(0,1) and strictly increasing if a>1a>1. For the canonical base ee, the inverse of the exponential e^{x}e^{x} is written \ln(x)\ln(x) and called the natural logarithm. The logarithmic functions have special properties which are a direct result of their definition as the inverses of exponential functions. If a\in (0,1)\cup (1,\infty )a\in (0,1)\cup (1,\infty ), with xx and y y positive and r\in \mathbb{ R}r\in \mathbb{ R}, then
(Log multiplication)\displaystyle \log_{a}(xy)=\log _{a}(x)+\log _{a}(y)\displaystyle \log_{a}(xy)=\log _{a}(x)+\log _{a}(y).
(Log division)\displaystyle \log_{a}\left(\frac{x}{y}\right)=\log _{a}(x)-\log _{a}(y)\displaystyle \log_{a}\left(\frac{x}{y}\right)=\log _{a}(x)-\log _{a}(y).
(Log exponent)\displaystyle \log_{a}(x^{r})=r\log _{a}(x)\displaystyle \log_{a}(x^{r})=r\log _{a}(x).
(Log base change)\displaystyle \log_a(x)=\frac{\log_b(x)}{\log_b(a)}\displaystyle \log_a(x)=\frac{\log_b(x)}{\log_b(a)}.
As a consequence of Log base change, we get a change of base formula for exponentials.
\displaystyle a^x=b^{\log_b (a^x)}=b^{x\log_b(a)}.
\displaystyle a^x=b^{\log_b (a^x)}=b^{x\log_b(a)}.
We derive formulas for the inverses of the hyperbolic functions using the formula for finding the inverse of a composite function:
$\newcommand{\T}{\rule{0pt}{3.0ex}} \newcommand{\B}{\rule[-2.2ex]{0pt}{0pt}} \begin{array}{|l|l|l|l|l|} \hline \textrm{Function} & \textrm{Definition} & \textrm{Domain} & \textrm{Range} \\ \hline \T\B\func{sinh}^{-1}x & \ln(x+\sqrt{x^2+1}) & \mathbb{R} & \mathbb{R} \\ \hline \T\B\func{cosh}^{-1}x & \ln(x+\sqrt{x^2-1})& [1,\infty ) & [0,\infty ) \\ \hline \T\B\func{tanh}^{-1}x & \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right) & (-1,1) & \mathbb{R} \\ \hline \T\B\func{coth}^{-1}x & \frac{1}{2}\ln\left(\frac{x+1}{x-1}\right)& x\notin[-1,1] & \mathbb{R}-\{0\} \\ \hline \T\B\func{sech}^{-1}x & \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right)& (0,1] & [0,\infty ) \\ \hline \T\B\func{csch}^{-1}x & \ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1}\right)& \mathbb{R}-\{0\} & \mathbb{R}-\{0\} \\ \hline \end{array}$