Can two blackbody emitters combine to act like one emitter?

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This discussion centers on the mathematical proof that two blackbody emitters can be represented as a single blackbody emitter. The equation under consideration is the sum of two functions, \(\frac{1}{e^{ax}-1} + \frac{1}{e^{bx}-1}\), which the user suspects can be expressed as \(\frac{1}{e^{cx}-1}\) for some function \(c\) derived from \(a\) and \(b\). The user graphed the functions and noted that while the shapes are similar, they are not identical, indicating that a straightforward combination may not exist. Further exploration suggested that the combined form might be \(\frac{k}{e^{cx}-1}\), where \(k\) is also a function of \(a\) and \(b\).

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I'm having a discussion about blackbody radiation on one of the other forums. I'm trying to prove that two blackbody emitters combine to act like one blackbody emitter. Unfortunately the algebra is beyond me. What I need to show is the following:

I want to show that
\frac{1}{e^{ax}-1}+\frac{1}{e^{bx}-1}

can be written as

\frac{1}{e^{cx}-1}

where c is some function of a and b. I suspect it might be a weighted average of a and b.

Can it be done? Having plotted a graph of the sum of the two functions I suspect it can since the shape of the graph is the same shape as the individual functions before they are added together.

Edit: Thinking about it it's probably more likely to be of the form

\frac{k}{e^{cx}-1}

where k is also a function of a and b. But I can't show that either. Putting things over a common denominator didn't help.
 
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The graphs may look "nearly" the same shape, but they are different.

If
y = 1 / ( e^ax - 1)
1/y = e^ax + 1
1/y - 1 = e^ax
ax = log(1/y + 1)

So the graph of log (1/y +1) against x is a strainght line.

Take say a = 1, b = 0.5, and x from 0.1 to 2.

Plot the corrsponding graph for the sum of the two functions, and it is close to a strainght line but not exactly straight.
 
Thanks AlephZero that helped a lot and saved me a lot of time I might have wasted trying to prove it was.
 

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