Question

How do you graph `y=log_4(x+1)`?

Answer 1

Change log base to `e` and scale `y = ln(x+1)` by `1/ln4`

Explanation:

`y=log_4 (x+1)`

Remember that: `log_b a = ln a/ln b`

Hence `y=ln(x+1)/ln4`

Therefore the graph of y will be the graph of `ln(x+1)` scaled by `1/ln4`

`ln(x+1)` is defined for `x> -1` and has a single zero at `x=0`

The graph of `y` is shown below:

graph{ln(x+1)/ln4 [-10, 10, -5, 5]}