# How do you graph  F(x)= log_(1/3)(x+5) ?

Nov 22, 2016

This is a decreasing log with a VA at $x = - 5$.

#### Explanation:

$f \left(x\right) = {\log}_{\frac{1}{3}} \left(x + 5\right)$

The vertical asymptote is found by setting $x + 5$ equal to zero.

$x + 5 = 0 \implies x = - 5$

The base of the log is $\frac{1}{3}$. A base that is less than one indicates that the graph is a decreasing log.

To find the $x$ intercept, set $f \left(x\right) = 0$

$0 = {\log}_{\frac{1}{3}} \left(x + 5\right)$

Rewrite as an exponential and solve.

${\left(\frac{1}{3}\right)}^{0} = x + 5$

$1 = x + 5$

$x = - 4$ when $f \left(x\right) = 0 \implies$ the $x$ intercept is $\left(- 4 , 0\right)$

The $y$ intercept is found by setting $x = 0$

$y = {\log}_{\frac{1}{3}} \left(0 + 5\right)$

$y = {\log}_{\frac{1}{3}} 5$

Use the change of base formula ${\log}_{a} b = \log \frac{b}{\log} a$

$y = \log \frac{5}{\log \left(\frac{1}{3}\right)} = - 1.46 \implies$the $y$ intercept is $\left(0 , - 1.46\right)$

See the graph below.

Alternatively, the function can be rewritten as an exponential and an $x y$ table can be constructed by choosing values of $y$ and finding corresponding $x$ values.

$y = {\log}_{\frac{1}{3}} \left(x + 5\right)$

$x + 5 = {\left(\frac{1}{3}\right)}^{y}$

$x = {\left(\frac{1}{3}\right)}^{y} - 5$

Then choose values of $y$ such as $- 2 , - 1 , 0 , 1 , 2$ and find the corresponding values of $x$. Plot the resulting $x y$ coordinates.