The exponential function
$f\left(x\right)={b}^{x}$ is one-to-one, with domain
$(\text{\u2212}\infty ,\infty )$ and range
$\left(0,\infty \right).$ Therefore, it has an inverse function, called the
logarithmic function with base$b.$ For any
$b>0,b\ne 1,$ the logarithmic function with base
b , denoted
${\text{log}}_{b},$ has domain
$(0,\infty )$ and range
$\left(\text{\u2212}\infty ,\infty \right),$ and satisfies
${\text{log}}_{b}\left(x\right)=y\phantom{\rule{0.2em}{0ex}}\text{if and only if}\phantom{\rule{0.2em}{0ex}}{b}^{y}=x.$
The most commonly used logarithmic function is the function
${\text{log}}_{e}.$ Since this function uses natural
$e$ as its base, it is called the
natural logarithm . Here we use the notation
$\text{ln}(x)$ or
$\text{ln}\phantom{\rule{0.1em}{0ex}}x$ to mean
${\text{log}}_{e}\left(x\right).$ For example,
and their graphs are symmetric about the line
$y=x$ (
[link] ).
At this
site you can see an example of a base-10 logarithmic scale.
In general, for any base
$b>0,b\ne 1,$ the function
$g\left(x\right)={\text{log}}_{b}(x)$ is symmetric about the line
$y=x$ with the function
$f\left(x\right)={b}^{x}.$ Using this fact and the graphs of the exponential functions, we graph functions
${\text{log}}_{b}$ for several values of
$b>1$ (
[link] ).
Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.
Rule: properties of logarithms
If
$a,b,c>0,b\ne 1,$ and
$r$ is any real number, then
Now we can solve the quadratic equation. Factoring this equation, we obtain
$\left({e}^{x}-3\right)\left({e}^{x}-2\right)=0.$
Therefore, the solutions satisfy
${e}^{x}=3$ and
${e}^{x}=2.$ Taking the natural logarithm of both sides gives us the solutions
$x=\text{ln}\phantom{\rule{0.1em}{0ex}}3,\text{ln}\phantom{\rule{0.1em}{0ex}}2.$
The solution is
$x={10}^{4\text{/}3}=10\sqrt[3]{10}.$
Using the power property of logarithmic functions, we can rewrite the equation as
$\text{ln}\left(2x\right)-\text{ln}\left({x}^{6}\right)=0.$ Using the quotient property, this becomes
$\text{ln}\left(\frac{2}{{x}^{5}}\right)=0.$
Therefore,
$2\text{/}{x}^{5}=1,$ which implies
$x=\sqrt[5]{2}.$ We should then check for any extraneous solutions.
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |.
The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.