How do i find the inverse of: $f(x) = {2^{x - 1}} - 3$ $f(x) = 2^{x - 1} - 3$
My approach:
Take logs to base 2:
$ = \log_2 \left( x - 1 \right) - \log_2 \left( 2^3 \right)$
$ = \log_2 \left( {x - 1} \over {2^3} \right)$
This isn't the answer in the book, which is:
$f^{-1}(x) = \log_2 (x + 3) + 1$
 A: Here's the classic "switch $x$ and $y$ approach."
$$x=2^{y-1}-3\\x+3=2^{y-1}\\y-1=\log_2(x+3)\\y=\log_2(x+3)+1$$ Alternately, if you'd prefer to avoid $y$: $$x=f\bigl(f^{-1}(x)\bigr)\\x=2^{f^{-1}(x)-1}-3\\x+3=2^{f^{-1}(x)-1}\\f^{-1}(x)-1=\log_2(x+3)\\f^{-1}(x)=\log_2(x+3)+1$$

You seem shaky with your log rules, so it might be better to avoid arithmetic manipulations altogether, here.
We can think of $f(x)$ as the graph of the function $x\mapsto 2^x$ shifted right by $1$ and down by $3$. We know how to invert $x\mapsto 2^x$--that's just $x\mapsto\log_2(x),$ but how do the shifts change things? To obtain the graph of an inverse function, we simply reflect the original function's graph over the line $y=x$. Hence, when we shifted to the right by $1$, our inverse shifted up by $1$, and when we shifted down by $3$, our inverse shifted left by $3$. Indeed, $f^{-1}(x)=\log_2(x+3)+1$ is simply the graph of $x\mapsto\log_2(x)$ shifted left by $3$ and up by $1$, as desired.
A: Your defining equation for the function is  $y = 2^{x-1} - 3 .$  This is no general rule for the logarithm of a difference of terms, so you must move the '3' first.  You then have
$y + 3  = 2^{x-1}$ .  You can now use logarithms from this point.  Once you have solved for $x$  , you can swap variables to write your expression as an inverse function on  $x$.  
A: You didn’t take logs to base $2$: it isn’t true that $\log_2(u-v)=\log_2 u-\log_2v$, and it isn’t true that $\log_22^{x-1}=\log_2(x-1)$. Moreover, taking the log base $2$ of $y=2^{x-1}-3$ doesn’t solve for $x$, which is what you need to do in order to find the inverse function.
What you want to do is solve $y=2^{x-1}-3$ for $x$. Add $3$ to both sides: $y+3=2^{x-1}$. Now take logs base $2$ to get $x-1$ out of the exponent: $\log_2(y+3)=\log_2 2^{x-1}=x-1$. Finally, add $1$ to both sides, and you have $x$: $x=1+\log_2(y+3)$. This tells you that $f^{-1}(y)=1+\log_2(y+3)$. You’re probably expected to express $f^{-1}$ with $x$ as the independent variable instead of $y$, so just replace $y$ everywhere that it occurs with $x$: $f^{-1}(x)=1+\log_2(x+3)$.
A: Finding the inverse of a function can be done by re-arranging to make $x$ the subject of the equation:
$$f(x)=2^{x-1}-3\implies2^{x-1}=f(x)+3 \implies x=\log_{2}\left(f(x)+3\right)+1$$
We now rename $x$ to $f^{-1}(x)$ and $f(x)$ to $x$ to get our inverse function:
$$f^{-1}:x\mapsto \log_{2}(x+3)+1 \quad\text{or}\quad f^{-1}(x)=\log_{2}(x+3)+1$$
As required.
