Precalculus (6th Edition) Blitzer

Step 1. Replace $f(x)$ with $y$; we have $y=2^x$ Step 2. Switch $x$ and $y$; we have $x=2^y$ Step 3. Solving for $y$, we have $y=log_2(x)$, with $x\gt0$ Step 4. Replace $y$ with $f^{-1}(x)$, we get $f^{-1}(x)=log_2(x)$ with $x\gt0$ Step 5. We can see that a possible problem is that step 3 requires an advanced concept plus a restriction to the values of $x$.