So I was trying to solve the following equation. I'm fairly good at mathematics so the fact that I have no idea what to do in order to solve this question kind of annoys me. I thought I'd see if anyone here can solve it.
(note) I'm looking for an exact answer.
$2^x \textrm{ln}(2)+3^x \textrm{ln}(3)-\textrm{ln}(324)=0$
On
Unfortunately, I don't believe that you can easily find an exact solution to this, and depending on where this problem came from, you may have to resort either to numerical methods or to showing only that a solution must exist, whatever it may be.
Here are some ways of rewriting the problem, which might show why it is difficult to solve. Notice for instance that you can rewrite it as a differential equation $$g'(x) = \ln(324),$$ where $g(x) = 2^x+3^x$. By integrating this amounts to solving an equation of the form $$g(x) = \ln(324)x+C,$$ i.e. $$2^x+3^x = \ln(324)x+C$$ for $x$. This is a harder equation to solve than the more harmless looking $2^x=x$, which in itself is not solvable in terms of the usual elementary functions.
Another approach is to rewrite your equation as $$\ln(2^{2^x})+\ln(3^{3^x}) = \ln(324),$$ or $$\ln(2^{2^x}\cdot3^{3^x}) = \ln(324).$$ Removing the logs, you get $$2^{2^x}\cdot3^{3^x}=324.$$ This is equivalent to the original equation, and might give you another look at what the nature of the problem is.
In the most general case, equations of the form $$f(x)=a_1\times b_1^x+a_2\times b_2^x+c=0$$ do not show explicit solutions and only numerical methods can be used.
The simplest root finding method is probably Newton which, starting with a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Let us consider the function $$f(x)=2^x \log(2)+3^x \log(3)-\log(324)$$ $$f'(x)=2^x \log ^2(2)+3^x \log ^2(3)$$ Looking at the graph of the function, there is a root close to $x=1$. So, let us start Newton procedure using $x_0=1$; the successive iterates will then be $1.239779907$, $1.214920613$, $1.214600154$, $1.214600102$ which is the solution for ten significant digits.
We could do better consider a transform which will make the function more linear (then better conditioned for the root-finding procedure). What I propose (plot the new function to see why) is to use $$g(x)=\log\Big(2^x \log(2)+3^x \log(3)\Big)-\log\Big(\log(324)\Big)$$ the graph of which being almost a straight line. The derivative is $$g'(x)=\frac{2^x \log ^2(2)+3^x \log ^2(3)}{2^x \log (2)+3^x \log (3)}$$
Starting again Newton with $x_0=1$, the successive iterates will now be $1.215396717$, $1.214600113$, $1.214600102$.
Ten significant digits in three iterations does not seem too bad (at least to me).