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compro21 · Jun 20, 2007, 09:04 PM

How would I go about solving an equation like this:
2^x+3^x=13 ?

The answer is trivial, x=2, but is there anyway to find it without guessing or using a finite differencing machine (i.e. computer approx.)? The function is onto and 1-to-1 which means there is only one, unique solution, but I have still not found any analytical way of solving it. Anyone else have any ideas?

Haxzor50 · Jun 20, 2007, 09:53 PM

Ahhh I hate functions and lines and...

Capuchin · Jun 21, 2007, 12:56 AM

You can use
$ln(a+c) = ln(a) + ln(1+e^{ln(c) - ln(a)})$

Capuchin · Jun 21, 2007, 04:37 AM

Actually, I think I am mistaken.

I think that this type of question is notoriously hard to solve (See Fermat's Last Theorem).

compro21 · Jun 21, 2007, 08:40 AM

Fermat's last theorem only applies to integer triples each raised to the same power, i.e. Pythagorean triples.

Has this problem been solved before?

galactus · Jun 21, 2007, 09:10 AM

There are probably numerous methods, but you could always try Newton's method.

We know x=2, but say we don't. Then try an initial guess of 1.5.

$1.5-\frac{2^{1.5}+3^{1.5}-13}{ln(2)\cdot{2^{1.5}}+ln(3)\cdot{3^{1.5}}}=2.148 76424149$

Perform the iterations and it converges to 2 rather quickly. If you use an initial guess closer to 2 it'll converge faster.

Just a thought.

ebaines · Jun 21, 2007, 10:43 AM

I don't believe that there is a closed-form solution for this type of problem, so the best you'll be able to do is using approximation techniques. However, after messing with this for a while I did find that it can be reworked to get the unknown out of the exponent. If we substitute for $x$ the variable $w$ defined by:

$ x = log_2(w) $

Then after some manipulation you can get to:

$ w + w^{log_2 3} = 13 $

Since $log_2 3$ is just a number, let's call it C, and the equation you then need to solve is:

$ w+w^c = 13 $

Somehow this seems to me like it ought to be easier to solve, since the variable you are trying to solve for is no longer in the exponent, but I can't come up with a closed-form method.

asterisk_man · Jun 23, 2007, 06:14 AM

I couldn't see a way to the solution and maxima doesn't find a solution either.
I agree with galactus, newton's method is probably best here. Also, make sure there's not a typo :)

compro21 · Jun 23, 2007, 10:47 AM

Typo somewhere?