Is there a way (excluding solving by graphing the two functions) to solve an equation like $\sin x = \ln x$ ?
Solving simple transcendental equations
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If you mean finding a closed form by mean of elementary functions: in this example, no. By elementary, I mean trig, log, exponential and their inverses, and of course the four operations and the extraction of roots (this last operation can be computed as $a^b=\exp(b\log a)$ anyway). Sometimes there is a solution, it depends entirely on the equation.
If you mean proving there is a solution: yes, it's possible, mainly by studying the variations of the functions.
If you mean finding an approximate solution: there are many ways. The most common is maybe Newton's method, at least for such a simple example.
Notice that finding a closed form is generally impossible. Often it's also not a good idea. For instance, it's relatively easy to solve a polynomial equation of degree $4$. However, the closed form is not necessarily well behaved numerically, and it may be wiser to use a numerical method if you want a numerical value, even though there is actually a "formula".
Notice also that the equation (or another equation with no closed form solution) may define a "special function", that you can then use as if it was a closed form solution. For instance, it's common to define the solution of $xe^x=a$ to be the Lambert $W$ function, so the solution is simply $x=W(a)$. But this seemingly closed form actually hides a computation that can be done only numerically. It's still interesting to do that because you can study the function $W$, its variations, find a Taylor series expansion, and many more.
So, what do you want to do exactly?
$$\sin x=\log x$$ $$x=e^{\sin x}$$ so $$x=e^{\sin e^{\sin e^{\sin e^{\sin .......}}}}=2.219...$$ or you can use Newton's Method $$x_n=x_{n+1}-\frac{y_n}{y'_n}$$ so that $$y=\sin x-\log x$$ $$y'=\cos x-\frac{1}{x}$$