Answer:
[tex]f^{-1}(x)=\frac{1}{2}ln(x)+2[/tex]
Step-by-step explanation:
Start by changing the f(x) into a y. Then switch the x and the y. Then solve for the new y. Like this:
[tex]y=e^{2x-4}[/tex] becomes
[tex]x=e^{2y-4}[/tex]
To solve for the new y, we need to get it out of its current exponential position which requires us to take the natural log of both sides. Since a natural log has a base of e, natural logs and e's "undo" each other, just like taking the square root of a squared number.
[tex]ln(x)=ln(e)^{2y-4}[/tex]
When the ln and the e cancel out we are left with
ln(x) = 2y - 4. Add 4 to both sides to get
ln(x) + 4 = 2y. Divide both sides by 2 to get
[tex]\frac{1}{2}ln(x) + 4 = y[/tex].
Since that is the inverse of y, we can change the y into inverse function notation:
[tex]f^{-1}(x)=\frac{1}{2}ln(x)+4[/tex]
