How to find inverse of...

Question

How do you find the inverse of the equation in the form $y= b^{x-h} +K$ for example: $y=2^{x-4} +6$ I already know that the inverse of $b^x$ is $\log_bx$ but how do you find with the $H$ and $K$ value included?

Answer 1

You just want to solve for $x$ in $$ Y = b^{x-h} + K. $$ You get $$\begin{align*} Y - K &= b^{x-h} \quad &\Rightarrow \\ \ln(Y-K) &= (x-h) \ln(b). \quad &\Rightarrow \\ &... \end{align*} $$ As you note in your question, you can of course use $\log_b$ instead of $\ln = \log_e$ if you prefer.

Answer 2

You can think of it in much of the same way. To find the inverse of the function $y=f(x)$ you write $x=f(y)$ and solve for $y$.

In this case you get $x=b^{y-h}+k$ move $k$ over and take the log base $b$ of both sides and you get $\log_b(x-k)=y-h$

Another way you can look at this is that you are shifting $x$ to the right by $h$ units and shifting $y$ up by $k$ units. The inverse of this must then be shifted right $k$ units and down by $h$ units.

Answer 3

For function $y=b^{x-h}+K$, the first part in obtaining the inverse function is to rearrange the equation solving for $x$:-

$b^{x-h}=y-K \Rightarrow(x-h)=\log_b(y-K)\Rightarrow x=h+\log_b(y-K)$

To obtain the inverse function you need to swap the $x$ and $y$:-

$y=h+\log_b(x-K)$