Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Do you have some assumptions on $a,b,k_1,k_2$? e.g. if $k_1 = k_2 = 0$ the function is constant. – Surb Jul 31 '14 at 23:46
For most $k_1, k_2$ you cannot do it in terms of elementary functions. – André Nicolas Jul 31 '14 at 23:47
@Surb Let's suppose all of them are positive numbers. – LCFactorization Jul 31 '14 at 23:55
up vote 1 down vote accepted

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 $|b|$.

share|cite|improve this answer
This is probably the best approach. But do you have more details simple to understand its implementation? – LCFactorization Aug 1 '14 at 2:11

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:


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

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

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

share|cite|improve this answer
What happened to $k_1$ and $k_2$ in the second row? – MathFacts Aug 1 '14 at 0:10
It is k_1=k_2 above – Marc Aug 1 '14 at 0:14

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.

share|cite|improve this answer
Hard to know what OP means by positive number. We can do it in cases closely related to the ones mentioned above, such as $k_2=2k_1$, like $1/3,2/3$. – André Nicolas Aug 1 '14 at 1:33
@AndréNicolas: Agreed – MPW Aug 1 '14 at 12:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.