How to solve for x??:
(2^x) + (3^x) = 13
Knowing there will only be one root from the properties of exponential functions, we could use a root finding technique, such as Newton's method, which converges to $\displaystyle x=2$ in 10 iterations on my calculator with an initial guess of $\displaystyle x_0=0$.
I would use inspection first to see if a rational root exists, but say the equation had instead been:
$\displaystyle 2^x+3^x=12$
Now, using Newton's method, we find:
$\displaystyle x\approx1.91768594489$
Have you studied differential calculus? Newton's method uses differentiation in its recursive approach. Newton's method essentially finds the x-intercept of the line tangent to the function for which we are trying to find a root at a value of x and uses that as a better approximation. Then this method is repeated recursively to converge to a root of the function. There are functions for which this doesn't always work for particular values of x, but for well-behaved functions like exponential functions, this is a good technique to use.
You have the correct derivative.
For the function:
$\displaystyle f(x)=2^x+3^x-k=0$
$\displaystyle f'(x)=\ln(2)2^x+\ln(3)3^x$ and so by Newton's method, we have the recursion:
$\displaystyle x_{n+1}=x_n-\frac{2^{x_n}+3^{x_n}-k}{\ln(2)2^{x_n}+\ln(3)3^{x_n}}$
As stated, when $\displaystyle k=13$ there certainly is no need for this technique, but I simply wanted to give a method for approximating the root when it is irrational, since you seemed to be interested in an "algebraic" method.