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?

Ahhh I hate functions and lines and...

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

Actually, I think I am mistaken.

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

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?

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.

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.

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 :)

Typo somewhere?