Inverse Function f(x) = ln(e^2x + e^x + 1) Solution

[Master J]

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

I want to find the inverse of this function.

I get:

e^y=e^2x + e^x + 1

If I say then that 1=e^0, then can I say y=2x + x? Therefore, x=y/3?

 

[Troels]

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

I want to find the inverse of this function.

I get:

e^y=e^2x + e^x + 1

If I say then that 1=e^0, then can I say y=2x + x? Therefore, x=y/3?

You got it right as far as taking the exponential on both sides of the equation. From there it helps to notice that:

$$e^{2x}=\left(e^x\right)^2$$

Thus converting the entire right hand side into a quadric equation:

$$e^{y}=u^2+u+1$$

where $$u=\exp(x)$$ and whilst treating $$\exp(y)$$ as a constant, solve this with respect to u

 

[HallsofIvy]

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

I want to find the inverse of this function.

I get:

e^y=e^2x + e^x + 1

If I say then that 1=e^0, then can I say y=2x + x? Therefore, x=y/3?

No, you can't. e^2x+ e^x+ e⁰ is NOT equal to e^{2x+ x+ 0}

As Troels said, you can replace e^x by u and get the equation e^y= u²+ u+ 1 or u²+ u+ (1- e^y)= 0 and solve that with the quadratic equation.

That gives
$$u= \frac{-1\pm\sqrt{4e^y- 3}}{2}$$
You might think that gives two solutions and so there is no "inverse" function (I did at first) but since u= e^x, u cannot be negative: we must take the positive root:
$$u= e^x= \frac{-1+ \sqrt{4e^y- 3}}{2}$$
so
$$x= ln(\frac{-1+ \sqrt{4e^y- 3}}{2})$$

and the inverse function is
$$f^{-1}(x)= ln(\frac{-1+ \sqrt{4e^x- 3}}{2})$$