Trying to find an inverse equation - maybe cosh/sinh

[meee]

Given $$f(t) =$$ $$\frac{2e^{t} + 3e^{-t}}{e^{t} + 2e^{-t}}$$ find $$f^{-1}(t)$$My attempt:
i worked for about half an hour but i think I am not doing the right thing. i tried multiplying it out, factorizing, pretty much just playing around with it.
OMMGG mayb got it

 

[HallsofIvy]

I would do this: write the equation as
$$x= \frac{2e^t+ 3e^{-t}}{e^t+ 2e^{-t}}$$
Now swap x and t:
$$t= \frac{2e^x+ 3e^{-x}}{e^x+ 2e^{-x}}$$
That "gives" the inverse function. Now you "only" have to solve for x!

Multiply that denominator on both sides:
$$te^x+ 2te^{-x}= 2e^x+ 3e^{-x}$$
Multiply the entire equation by e^x
$$te^{2x}+ 2t= 2e^{2x}+ 3$$
Let y= e^x so that you get a quadratic in y
$$ty^2+ 2t= 2y^2+ 3$$
Solve that by a simple square root, then take ln of both sides to find x as a function of x.

 

[meee]

ooooohh thankszzz
sooo $$x = ln\sqrt{\frac{3-2t}{t-2}}$$