Determine the inverse function of $f(x)=3^{x-1}-2$ Determine the inverse function of $$f(x)=3^{x-1}-2.$$ I'm confused when you solve for the inverse you solve for $x$ instead of $y$ 
so would it be $x=3^{y-1}-2$?

 A: $$
y=3^{x-1}-2
$$
$$
y+2=3^{x-1}/\cdot\log_3
$$
$$
\log_3(y+2)=\log_3 3^{(x-1)}
$$
$$
\log_3(y+2)=x-1
$$
$$
x=\log_3 (y+2)+1
$$
Now we take the substituon $x=y^{-1}$, and we have
$$
y^{-1}=\log_3 (x+2)+1
$$
A: In case of inverse you find $x$ in terms of y.
First you note that the inverse exists(monotonic function) then you do the following.
$y=3^{x-1}-2\Rightarrow y+2=3^{x-1}\Rightarrow \log_3(y+2)=x-1\Rightarrow x=\log_3(y+2)+1$
And you are done.(As a customary one needs to replace x with y as we generally denote a function by y with independent variable x)
A: Yes, you are right till here. Recall that you have to solve for $y$ and finally replace $y $ with $f^{-1}(x)$ and $x$ with $y$.
$$x = 3^{y-1} - 2$$
Just add $2$ and take log base 3 on both sides.$$\begin{align} &x + 2 = 3^{y-1} \\  \Rightarrow & \log_3(x + 2) = y - 1 \\ \Rightarrow & y = \log_3(x + 2) + 1  \end{align}$$And you have $f^{-1}(x) = \log_3(x + 2) + 1$. Does it make sense now?
A: To find the inverse function $f^{-1}$ in the form $y = f^{-1}(x)$, there are two things you can do:
1) Switch $x$ and $y$, and then solve for $y$.
2) Solve for $x$, and then switch $x$ and $y$.
You seem to be trying to switch $x$ and $y$ and then solve for $x$.  You need to solve for $y$ if you've already switched the variables.  See the methods (1) and (2) above.
A: To find the inverse start by writing $y=3^{x-1}-2$. Now, can you rearrange it so you get $x=\ldots$ ?
For example $y+2 = 3^{x-1}$. Then how do you "unpick" a power? You take the logarithm:
\begin{array}{ccc}
y+2 &=& 3^{x-1} \\ \log_3(y+2) &=& \log_3(3^{x-1}) \\ \log_3(y+2) &=& x-1
\end{array}
Finally, $x=\log_3(y+2)+1$. Your inverse function is then $\operatorname{f}^{-1}(x) = \log_3(x+2)+1$.
A: it's right, but I think you need simplify it. $y=log_3(x+2)+1$
