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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?

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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 at 23:46
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For most $k_1, k_2$ you cannot do it in terms of elementary functions. –  André Nicolas Jul 31 at 23:47
    
@Surb Let's suppose all of them are positive numbers. –  LCFactorization Jul 31 at 23:55

3 Answers 3

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 $|b|$.

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This is probably the best approach. But do you have more details simple to understand its implementation? –  LCFactorization Aug 1 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:

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

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

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

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

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What happened to $k_1$ and $k_2$ in the second row? –  MathFacts Aug 1 at 0:10
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It is k_1=k_2 above –  Marc Aug 1 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.

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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 at 1:33
    
@AndréNicolas: Agreed –  MPW Aug 1 at 12:09

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