Question

How do you find the inverse of `y=log_x4`?

Answer 1

`y=4^{1/x}`

Explanation:

I would recommend using the base change rule for this one.

`log_ba=log_ca/log_cb`

Let's express `log_x4` as `ln4/lnx`

(Note: As an IB student I use `ln` to denote log base e (`log_ex`). I am aware that there may be different conventions regarding this depending on region/educational system etc.)

Therefore `y = ln4/lnx`

We rearrange the equation to isolate x.

`ln4/y = lnx`

Rewrite the equation using e raised to the power of both sides so that the equation is equal to x.

`e^{ln4/y} = e^{lnx}=x`

Using the relation `a^{x} = e^{xlna}` we can write `e^{ln4/y}` as `4^{1/y}`.

This gives us `x=4^{1/y}`

Having rearranged the equation all we need to do is swap y and x to find the inverse.

`y=4^{1/x}`