Inverse function of bi-exponential function?

AI Thread Summary
The discussion focuses on finding the inverse function of a bi-exponential function, specifically of the form Y=A*exp(B*X)+C*exp(D*X). It is noted that while single exponential functions can be inverted using logarithms, bi-exponential functions present challenges, particularly due to their potential lack of a straightforward inverse. The bi-exponential decay equation discussed has parameters that must meet specific conditions for the function to be valid. It is suggested that restricting the domain may allow for an inverse, but generally, graphical or numerical methods are recommended for finding approximate solutions. Overall, the complexity of the bi-exponential function limits the ability to derive a simple inverse analytically.
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Homework Statement


What is the inverse function of a bi-exponential function like the following:
Y=A*exp(B*X)+C*exp(D*X)

Homework Equations


Y=A*exp(B*X)


The Attempt at a Solution


If it is a single exponential function, i can take log on both sides to get inverse function. But when it comes to a bi-exponential function, i really don't know how to do it. Thanks in advance.
 
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If y(x)=(1/2)(e^x+e^(-x)) (that's the cosh function), then y(1)=y(-1). It doesn't even have an inverse.
 
In fact, the bi-exponential equation i described here is a bi-exponential decay equation. The domain of X is 0 to inf. The parameter A and C should be greater than 0 and B, D should be less than 0. I have plotted a simulated data with R. It seems the inverse function of the bi-exponential function should be a multiple logarithmic function.
 
Ok, if you restrict the domain, then it can have an inverse. But I don't think there is a nice way to solve for x in terms of y except in special cases. You'll have to use graphical or numerical techiques to get an approximate solution. As you've noticed, logs don't get you there.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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