Find the inverse function of f(x)=ln3x^e+5?

Answer 1

I think he did make a mistake and got ln(x^e) mixed up with ln(e^x).

ln(e^x) = x, but ln(x^e) = eln(x) which does not equal x.

If the function is f(x) = ln(3x^e) + 5,

then replace f(x) with y, for simplicity's sake.
y = ln(3x^e) + 5, then swap x and y.

x = ln(3y^e) + 5, then solve for y...subtract 5 from both sides

x - 5 = ln(3y^e), then exponentiate both sides

e^(x - 5) = e^(ln(3y^e)) simplify the right side

e^(x - 5) = 3y^e, divide both sides by 3

(e^(x - 5))/3 = y^e, take the e-root of both sides
((e^(x - 5))/3)^(1/e) = y = finv(x)

Yuck!

Answer 2

envidiar's answer is clearly correct. However, I do not understand step 7: the right side of the equation is ln(x^e), which is e*ln(x). How does that become x?

Edit:

In response to later answers, envidiar assumed the equation was f(x)=ln[3x^e + 5], with the 5 inside the logarithm. If you take his answer and use it as f*(f(x)), the result should be x (the inverse function of the original function is x). If you do that, you indeed get the correct result.

s_h_mc's Q&A: If I try finv(f(x)) in your result I do not get x
Eric S : Your solution works for your interpretation, but that does not solve the problem I have with enviear's solution (with +5 inside the logarithm).

Edit again: Applying Eric S method to the problem as interpreted by envidiar gives the solution:
finv(x) = [(e^x - 5)/3]^(1/e) which also returns the correct result for finv(f(x)) = x.

Answer 3

First off, the way you state your question is a bit unclear. So I'll assume the question is find the inverse of f(x)=ln(3x^(e)+5).
If the equation is different, make another question.
1) f(x)=ln(3x^(e)+5).

2) y=ln(3x^(e)+5).

3) e^(y)=3x^(e)+5; e both sides and get this

4) e^(y)-5=3x^(e); minus the 5 to the other side

5) (e^(y)-5)/3=x^(e); divide the other side by 3

6) ln((e^(y)-5)/3)=ln(x^(e)); ln both sides

7) ln((e^(y)-5)/3)=x; left side remains the same, e is brought out in front of ln and cancel each other out, leaving only x

8) y^(-1)=ln((e^(y)-5)/3); change x to y, and y to x, or call it y^(-1)

Answer 4

f(x) = y = (ln3) * (x^e) + 5

Switch variables, and solve:

x = (ln3) * (y^e) + 5
x - 5 = (ln3) * (y^e)
(x - 5) / (ln3) = y^e

ln [ (x - 5) / (ln3) ] = ln y^e

ln [ (x - 5) / (ln3) ] = e * ln y
ln [ (x - 5) / (ln3) ] / e = ln y

y = e^ { (ln [ (x - 5) / (ln3)]^[1/e] )}
y = [(x - 5) / (ln3)] ^ [1/e]