Inverse of a logrithmic function

[Timcago]

Find the inverse of the function f(x)=3^(x-2)+1. What is the domain of f^-1

Here is my progress:

1. change f(x) to y

y=3^(x-2)+1

2. Switch x and y

x=3^(y-2)+1

3. Solve for y


Here is where i am stuck. My theory is that i need to take the natural log of both sides of the equation, but the +1 really complicates the situation.

ln(x)=ln(3^(y-2))+ln(1)

4. now according to a rule the right side of the equation becomes the power it is raised to?

ln(x)=y-2+ln(1)

5. subtract ln(1) and add 2 to the other side?

ln(x)-ln(1)+2=y

y=ln(x)+2

6. now change to inverse form

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

The domain is (0,+(infinity))

Am i right? If i am not where did i go wrong?

 

[JakeD]

Find the inverse of the function f(x)=3^(x-2)+1. What is the domain of f^-1

Here is my progress:

1. change f(x) to y

y=3^(x-2)+1

2. Switch x and y

x=3^(y-2)+1

3. Solve for y


Here is where i am stuck. My theory is that i need to take the natural log of both sides of the equation, but the +1 really complicates the situation.

ln(x)=ln(3^(y-2))+ln(1) Here's a problem. Logs don't work across a sum.

4. now according to a rule the right side of the equation becomes the power it is raised to?

ln(x)=y-2+ln(1) This isn't correct either.

\(\displaystyle \Large \begin{array}{ll}
x &= 3^{y-2} + 1 \\
x -1 &= 3^{y-2} \\
\ln(x-1) &= (y-2) \ln(3) \\
y -2 &= \ln(x-1) / \ln(3) \\
y &= \ln(x-1) / \ln(3) + 2 .
\end{array}\)

You can check this by plugging back into the original function:

\(\displaystyle \Large \begin{array}{ll}
f(y) &= f(\ln(x-1) / \ln(3) + 2) \\
&= 3^{(\ln(x-1) / \ln(3) + 2 -2)}+1 \\
&= 3^{(\ln(x-1) / \ln(3))}+1 \\
&= e^{\ln(3)(\ln(x-1) / \ln(3))}+1 \\
&= e^{\ln(x-1)}+1 \\
&= (x-1)+1 \\
&= x \\
\end{array}\)

as required by definition of an inverse.