Question

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

Answer 1

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

Explanation:

`f(x)=log_(1/3) (x+5)`

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

`x+5=0 => x=-5`

The base of the log is `1/3`. A base that is less than one indicates that the graph is a decreasing log.

To find the `x` intercept, set `f(x)=0`

`0=log_(1/3) (x+5)`

Rewrite as an exponential and solve.

`(1/3) ^0 = x +5`

`1=x+5`

`x=-4` when `f(x)=0 =>` the `x` intercept is `(-4,0)`

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

`y=log_(1/3) (0+5)`

`y=log_(1/3) 5`

Use the change of base formula `log_a b = logb/loga`

`y=log5/(log (1/3))=-1.46 =>`the `y` intercept is `(0, -1.46)`

See the graph below.


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

`y=log_(1/3) (x+5)`

`x+5 = (1/3)^y`

`x=(1/3)^y -5`

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