# Math 22 Logarithmic Functions

## Logarithm Function

 The logarithm ${\displaystyle \log _{a}x}$ is defined as
${\displaystyle \log _{a}x=b}$ if and only if ${\displaystyle a^{b}=x}$


## Definition of the Natural Logarithmic Function

 The natural logarithmic function, denoted by ${\displaystyle \ln x}$, is defined as
${\displaystyle \ln x=b}$ if and only if ${\displaystyle e^{b}=x}$


## Properties of the Natural Logarithmic Function

 Let ${\displaystyle g(x)=\ln x}$
1. The domain of ${\displaystyle g(x)}$ is ${\displaystyle (0,\infty )}$ and the range of ${\displaystyle g(x)}$ is ${\displaystyle (-\infty ,\infty )}$
2. The x-intercept of the graph of ${\displaystyle g(x)}$ is ${\displaystyle (1,0)}$
3. The function ${\displaystyle g(x)}$ is continuous, increasing, and one-to-one.
4. ${\displaystyle \lim _{x\to 0^{+}}g(x)=-\infty }$ and ${\displaystyle \lim _{x\to \infty }g(x)=\infty }$


## Inverse Properties of Logarithms and Exponents

 1.${\displaystyle \ln e^{\sqrt {2}}}$

2.${\displaystyle e^{\ln x}=x}$

3.${\displaystyle \ln {xy}=\ln {x}+\ln {y}}$

4.${\displaystyle \ln {\frac {x}{y}}=\ln x-\ln y}$

5.${\displaystyle \ln {x^{n}}=n\ln x}$


Exercises 1 Use the properties of logarithms to rewrite the expression as the logarithm of a single quantity

a) ${\displaystyle \ln(x-2)-\ln(x+2)}$

Solution:
${\displaystyle \ln(x-2)-\ln(x+2)=\ln {\frac {x-2}{x+2}}}$

b) ${\displaystyle 5\ln(x-6)+{\frac {1}{2}}\ln(5x+1)}$

Solution:
${\displaystyle 5\ln(x-6)+{\frac {1}{2}}\ln(5x+1)=\ln(x-6)^{5}+\ln[(5x+1)^{\frac {1}{2}}]=\ln[(x-6)^{5}{\sqrt {5x+1}}]}$

c) ${\displaystyle 3\ln x+2\ln y-4\ln z}$

Solution:
${\displaystyle \ln x^{3}+\ln y^{2}-\ln z^{4}=\ln {\frac {x^{3}y^{2}}{z^{4}}}}$

d) ${\displaystyle 7\ln(5x+4)-{\frac {3}{2}}\ln(x-9)}$

Solution:
${\displaystyle 7\ln(5x+4)-{\frac {3}{2}}\ln(x-9)=\ln(5x+4)^{7}-\ln(x-9)^{\frac {3}{2}}=\ln {\frac {(5x+4)^{7}}{(x-9)^{\frac {3}{2}}}}}$

Exercises 2 Solve for x.

a) ${\displaystyle \ln(2x)=5}$

Solution:
${\displaystyle \ln(2x)=5}$, so ${\displaystyle e^{5}=2x}$, hence ${\displaystyle x={\frac {e^{5}}{2}}}$

b) ${\displaystyle 5\ln x=3}$

Solution:
${\displaystyle 5\ln x=3}$, so ${\displaystyle ln{x^{5}}=3}$, so ${\displaystyle e^{3}=x^{5}}$, hence ${\displaystyle x={\sqrt[{5}]{e^{3}}}}$