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I'm wondering if there is a way to solve for x given the following equation:

$$A^x + B^x = C^x$$

where A, B, and C are known constants. For pythagorean triples, $x = 2$. I've seen a lot of stuff for sums of exponents (often using Taylor expansion) but I'm wondering if fixing it to 2 or 3 terms provides any shortcuts?

Another way to represent the problem could be finding the zeroes of the function

$$f(x) = A^x + B^x + C^x$$

WolframAlpha is somehow able to solve it but I'm not sure exactly what they are doing.

Martin Sleziak
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Matt Dodge
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  • For your expression $f(x)=$ you need a minus sign on $C^x$ – Ross Millikan Apr 01 '13 at 20:39
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    Wolfram Alpha uses some sort of numerical method. There is no "algebraic" solution except for limited cases. – vonbrand Apr 01 '13 at 20:44
  • @RossMillikan Yeah I realize the C in the second isn't the same as the C in the first, it has the opposite sign. They are still both constants though so the expression for x should be the same – Matt Dodge Apr 01 '13 at 21:05
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    @mattedgod you're wrong. $A^x+B^x+C^x$ doesn't have zeros. – Kaster Apr 01 '13 at 21:13
  • @Kaster wouldn't that depend on the values of A, B, and C? – Matt Dodge Apr 01 '13 at 21:24
  • @mattedgod have you ever seen negative value for exponential function? – Kaster Apr 02 '13 at 00:03
  • @Kaster No but I haven't really seen that many exponential functions :) That does make me curious about how you would go about solving that though, surely $(-1)^x$ has solutions, even if they are imaginary, right? – Matt Dodge Apr 02 '13 at 05:21
  • @mattedgod Please, show at least one root of the equation $(-1)^x = 0$. – Kaster Apr 02 '13 at 07:21
  • Tag ([tag:exponential-sum]) is not good choice for this question, as you can see if you read the tag-excerpt of tag-wiki. But I was not able to think of a better tag than ([tag:exponentiation]). – Martin Sleziak Jun 10 '13 at 12:12

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Alpha is doing a numeric solution to the equation. You can tell because the answer comes out as a decimal, not as some expression in roots. Techniques for this are discussed in any numerical analysis book. I like chapter 9 of Numerical Recipes-old versions are free online, but any other text will have it as well.

Ross Millikan
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  • Ah I figured this was what they were doing. The expression seems so simple it would seem like it would have a formulaic solution but looks can be deceiving. – Matt Dodge Apr 01 '13 at 21:06