Finding the inverse of an exponential function

Question

I have this function:

$$F_X(x) = \frac{3e^{2x}}{4} + \frac{3e^{4x}}{8} - 0.1$$

of which I am trying to find the inverse function, as in $u = F_X^{-1}(x)$.

I made it to this form:

$$u= F_X(x) = \frac{3}{4}\cdot e^{2x} \cdot [\frac{2+e^{2x}}{2}] - 0.1 $$

Then applied $\ln(x)$ on both sides to get:

$$\ln(u+0.1) = \ln(\frac{3}{4}) + 2x + \ln(2+e^{2x}) -\ln(2)$$

But I can not seem to express $\ln(2+e^{2x})$ in the form of $x$ so I can continue.

Any advice will be much appreciated

Answer 1

The problem is that your function is not invertible on $\mathbb R$. Note that $$F_X(\ln(\frac z2)) = \frac34 z + \frac38 z^2 - 0.1$$ Quadratic polynomials are never invertible on $\mathbb R$.
You can still find partial inverses by solving $$z^2 + 2z - \frac{4}{15}-y = 0$$ for $z$. The result will be a multivalued function of $y$. For each branch, $g_{1,2}(y)$, a partial inverse of $F_X$ is given by $$F_{X,i}^{-1}(y) = g_i(\ln \frac y2)$$ If you actually do the math, you will get $$g_{1,2}(y) = -1 \pm \sqrt{\frac{19}{15} + y}$$ And thus the two partial inverses $$F_{X,1}^{-1}(y) = \sqrt{\frac{19}{15} + \ln \frac y2} - 1\\ F_{X,2}^{-1}(y) = -\sqrt{\frac{19}{15} + \ln \frac y2} - 1$$

Answer 2

Let $e^{2x}=t$, $e^{4x}=t^2$, and get a quadratic equation for $t$.

Answer 3

Hint: $$m=e^{2x}$$ and Use:

$$(e^{2x}+1)^2=e^{4x}+2e^{2x}+1$$