Say I want to know if $f(x)$ is convex. Can I apply any convex function, strictly increasing function to it and preserve convexity?
Say $f(x),g(x)$ are convex and strictly positive and I want to know if $f(x)g(x)$ is convex, then can I say
$-log(f(x)g(x)) = -log(f)-log(g)$
Which would mean ANY product of strictly positive convex functions are also convex. This seems a bit too simple so I guess I'm missing something.
As already noted in the comments, if $f$ is convex and increasing and $g$ is convex, then $f \circ g$ is convex. If both $f$ and $g$ are $\mathcal{C}^2$, you can easily show this via derivatives: $$ {(f \circ g)}'' = {((f' \circ g)g')}' = (f'' \circ g) {(g')}^2 + f'(g) g'' \geq 0 $$ if $f'', f', g'' \geq 0$. The proof can be given also for functions which are just continuous.
However the converse is not true at all, so from the convexity of $f \circ g$ you cannot deduce anything on the convexity of $g$. For instance, take $f(x) = x^3$ and $g(x) = \sqrt{x}$ as functions of $\mathbb{R}^+$. $g$ is not convex, but $f$ is convex and increasing and $f \circ g(x) = x^{\frac{3}{2}}$ is strictly convex. This test can help you only if $f$ is linear: then convexity is preserved in both directions (or, if $f$ is decreasing, is swapped with concavity).
Moreover, in general there is no hope that the product $fg$ is convex if $f$ and $g$ are convex. Take for example $f(x) = x^2$ and $g(x) = {(x-1)}^2$. You need additional conditions, for example it is sufficient that $f$ and $g$ are both positive and the derivatives $f'$ and $g'$ have pointwise the same sign. Again, it is useful to think to the differentiable case and noting the identity $$ {(fg)}'' = f''g + g''f + 2f'g'. $$
