Exponents and Logarithms

Rule	Statement
Product of powers	$a^r\cdot a^s=a^{r+s}$
Power of a power	$(a^r{)}^s=a^{rs}$
Logarithm definition	${\log }_{b}x=r⟺b^r=x$
Natural log	$\ln x={\log }_{e}x$
Change of base	${\log }_{b}x={\displaystyle \frac{\ln x}{\ln b}}$

Rules for Working with Exponents

Rule	Example or Explanation
$\left(a^r\right)\left(a^s\right)=a^{r+s}$	$(a^2)(a^3)=(a\cdot a)(a\cdot a\cdot a)=a^5$
${\left(a^r\right)}^s=a^{rs}$	${\left(a^2\right)}^3=a^2\cdot a^2\cdot a^2=a^6$
$(ab{)}^r=a^r{b}^r$	$(a\cdot b{)}^2=(a\cdot b)(a\cdot b)=a^2\cdot b^2$
${\displaystyle \frac{a^s}{a^r}}=a^{s-r}$	${\displaystyle \frac{a^5}{a^2}}={\displaystyle \frac{a\cdot a\cdot a\cdot a\cdot a}{a\cdot a}}=a^3$
$a^0=1$ (if $a\ne 0$)	$(a^r)(a^0)=a^{0+r}\Rightarrow a^0=1$
$a^{-r}={\displaystyle \frac{1}{a^r}}$	${\displaystyle \frac{1}{a^r}}={\displaystyle \frac{a^0}{a^r}}=a^{0-r}=a^{-r}$
$a^{m/n}=\sqrt[n]{a^m}$ ($m,n$ integers)	$a^{1/2}=\sqrt{a},a^{-3/2}={\displaystyle \frac{1}{\sqrt{a^3}}}$

Logarithms

Suppose and $b\ne 1$. If $x=b^r$, then $r$ is said to be the logarithm of $x$ to the base $b$, written:

$$r={\log }_{b}x.$$

For example:

$$2^3=8⟺{\log }_{2}8=3$$$${10}^{-4}=0.0001⟺{\log }_{10}0.0001=-4.$$

• Because $b$ is positive, for any real number $r$. If , the expression ${\log }_{b}x$ is undefined.
• The logarithm with base $e=2.718281828\cdots$ (Euler's number) is called the natural logarithm, written $\ln x$: $$y=\ln x⟺x=e^y.$$ The function $e^x$ is also written $\exp (x)$.
• Change of base formula: For any positive $b$ ($b\ne 1$): $${\log }_{b}x=\frac{\ln x}{\ln b}.$$

Rules for Working with Logarithms

Let . Then:

• $\ln (ab)=\ln a+\ln b$
• $\ln (a^r)=r\ln a$
• $\ln \sqrt[n]{a}=\ln a^{1/n}={\displaystyle \frac{1}{n}}\ln a$
• $\ln {\displaystyle \frac{a}{b}}=\ln (a{b}^{-1})=\ln a-\ln b$
• $\ln 1=0$

Frequently Asked Questions

What makes the natural logarithm "natural"?

The base $e\approx 2.71828$ arises naturally in calculus: it is the unique base for which the derivative of $b^x$ equals $b^x$ itself (with no extra constant factor). Because of this, $\frac{d}{dx}{e}^x=e^x$ and $\frac{d}{dx}\ln x=\frac{1}{x}$, which are the cleanest possible forms. Any other base introduces a constant factor of $\ln b$.

Why can't you take the logarithm of a negative number?

Because for any real $r$ and any positive base $b$. The output of an exponential function is always positive, so its inverse (the logarithm) can only accept positive inputs. The logarithm of a negative number would require complex numbers.

How do I use logarithm rules to simplify $\ln \frac{x^3\sqrt{x+1}}{e^{2x}}$?

Apply the rules step by step: $$\ln \frac{x^3\sqrt{x+1}}{e^{2x}}=\ln (x^3)+\ln (\sqrt{x+1})-\ln (e^{2x})=3\ln x+\frac{1}{2}\ln (x+1)-2x.$$