Question

Let `f(x) = klog_2x` Given that `f^-1 (1) = 8` , what is the value of k?

Answer 1

`k=1/3`

Explanation:

Given `f(x)=klog_2x` and `f^-1(1)=8`

We know that, if `f^-1(x)=y` then `f(y)=x`.
So, in the second equation, this means that `f(8)=1`
We have the first equation there, so we substitute `x=8` and `f(x)=1` to get
`1=klog_2(8)`

I'm sure you know what to do from here to get the above answer.

Hint:- `log_xy^z=zlog_xy`
`log_x(x)=1`