How to find the inverse function in explicit form?

Question

For a function below:

$$f(x)=a\cdot e^{-k_1 x}+b\cdot e^{-k_2 x}$$

How can I obtain its inverse function in explicit form?

Answer 1

Writing the equation as $t + b t^{p} = y$, there is a series solution in powers of $b$:

$$ t = y + \sum_{j=1}^\infty (-1)^j \left(\prod_{i=0}^{j-2} (jp-i)\right) \dfrac{y^{j(p-1)+1} b^j}{j!}$$ convergent for small $|b|$.

Answer 2

The only case where you can (in general) find the inverse function is if $k_1=k_2\neq 0$ $x>0$ and $a+b \geq0$

We have then:

$x=ae^{-kx}+be^{-kx}$

$\iff ln(x)=ln(ae^{-ky}+be^{-ky})=ln((a+b)(e^{-ky}))=ln(a+b)+ln(e^{-ky})=ln(a+b)-ky$

$\iff ky=ln(a+b)-ln(x) \iff y=\frac{ln(a+b)-ln(x)}{k}$

Well, I think thats still not what you are looking for. Maybe someone got other ideas?

Answer 3

This is equivalent to being able to find a local inverse of the function $$g(x) =ax^{k_1}+bx^{k_2}$$

because then $$f(x) =g(e^{-x})$$

so $$f^{-1}(y)=-\log g^{-1}(y)$$

You can do this explicitly if each $k_j$ is a nonnegative integer no greater than $4$, but higher values must be considered case by case.