Answer:
[tex]f^{-1}(x)=\frac{1}{6}\ln(x+9)[/tex]
Step-by-step explanation:
So we have the function:
[tex]f(x)=e^{6x}-9[/tex]
To solve for the inverse of a function, change f(x) and x, change the f(x) to f⁻¹(x), and solve for it. Therefore:
[tex]x=e^{6f^{-1}(x)}-9[/tex]
Add 9 to both sides:
[tex]x+9=e^{6f^{-1}(x)}[/tex]
Take the natural log of both sides:
[tex]\ln(x+9)=\ln(e^{6f^{-1}(x)})[/tex]
The right side cancels:
[tex]\ln(x+9)=6f^{-1}(x)[/tex]
Divide both sides by 6:
[tex]f^{-1}(x)=\frac{1}{6}\ln(x+9)[/tex]
And we're done!