3.7 Logarithmic Functions

A logarithmic function is a function of the form

$f(x)={\log }_{\mathrm{a}}x$

where $a$ is a positive real number, $a>0$, and $a\ne 1$. The domain of $f$ is the set of all real numbers $x$ such that $x>0$. While we may not be able to find the exact solution of a logarithmic function given a certain $x$ without a calculator’s help, we can apply the properties we learned in the previous section to find the domain of the function.

Suppose there are three logarithmic functions:

$f(x)=\ln (-2x+3)$

$g(x)={\log }_4(x^2+3)$

$h(x)=\ln \left(\frac{-4x+7}{5x+2}\right)$

Let’s start with the first function. It cannot be simplified any more. We know that the domain of the function is $-2x+3>0$, so $x<\frac{3}{2}$ .

For the function $g(x)$, even though we do not have to do anything to find its domain, let’s use one of the properties of logarithms. We will change the base from 4 to 2:

$g(x)={\log }_4(x^2+3)$

$=\frac{{\log }_2(x^2+3)}{{\log }_2(4)}$

$=\frac{{\log }_2(x^2+3)}{2}$

Thus, we can verify that the domain of this function is all real numbers.

Let’s look at $x^2+3$. Since the coefficient of $x^2$ is positive, we know that if there are two solutions, there will be some values of $x$ that make x^2+3 negative. Since

$b^2-4ac=0^2-4\cdot 1\cdot 3=-12<0$

this function has no solution. Therefore, $g(x)>0$ for any real number of $x$.

Finally, let’s look at the last function. Since it is a logarithm of division (even though we do not have to apply the property),

$\begin{array}{rl}h(x)& =\ln \left(\frac{-4x+7}{5x+2}\right)\\ & =\ln (-4x+7)-\ln (5x+2)\end{array}$

Since $-4x+7>0$ and $5x+2>0$, $x$ has to satisfy $x<\frac{7}{4}$  and $x>-\frac{2}{5}$ , so the domain of the function is $-\frac{2}{5}<x<\frac{7}{4}$ .