Hey I am having some trouble finding the inverse function of

f(x) = (2^x)-1

and also I need help with proving the functions are mutually inverse can you please show me how to do it.

\[ \text{If }y = \left(2^x\right)^{-1}\\ \text{then for the inverse, after swapping }x\text{ and }y, \]

\begin{align*} x &= \left(2^y\right)^{-1}\\ \frac{1}{x} &= 2^y\\ \log_2 \frac1x &= y \end{align*}

\[ \therefore f^{-1}(x) = \log_2 \frac{1}{y} \]

Ensure that you

**know** that the exponential and logarithm are inverse functions. Because the logarithm is

**defined** to be the inverse function to the exponential, the statements \(a^x = b\) and \(x = \log_a b\) are equivalent. i.e. we can go between one another.

Also, because they are inverses, the mutual inverses property is

**assumed** to hold here. That is, we can

**always** assume that \( a^{\log_a x} = x\) for \(x > 0\), and \(\log_a (a^x) = x\) for all real \(x\).