3.7 Logarithmic Functions

A logarithmic function is a function of the form

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

where $a$ $a$ is a positive real number, $a > 0$ $a > 0$ , and $a \neq 1$ $a \neq 1$ . The domain of $f$ $f$ is the set of all real numbers $x$ $x$ such that $x > 0$ $x > 0$ . While we may not be able to find the exact solution of a logarithmic function given a certain $x$ $x$ 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:

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

$g (x) = {log}_{4} (x^{2} + 3)$ $g (x) = \log_{4} (x^{2} + 3)$

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

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

For the function $g (x)$ $g (x)$ , 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:

$g (x) = {log}_{4} (x^{2} + 3)$ $g (x) = \log_{4} (x^{2} + 3)$

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

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

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

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

$b^{2} - 4 a c = 0^{2} - 4 \cdot 1 \cdot 3 = - 12 < 0$ $b^{2} - 4 a c = 0^{2} - 4 \cdot 1 \cdot 3 = - 12 < 0$

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

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),

$\begin{matrix} h (x) & = ln (\frac{- 4 x + 7}{5 x + 2}) = ln (- 4 x + 7) - ln (5 x + 2) \end{matrix}$ $\begin{aligned} h (x) & = \ln (\frac{- 4 x + 7}{5 x + 2}) \\ = \ln (- 4 x + 7) - \ln (5 x + 2) \end{aligned}$

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

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