I have two exponential equation problems. Firstly I have the problem $15^{−x^3+x^2−2} =15$, which I have simplified to $-x^3 + x^2 -2 = 1$. I am pretty sure this can't be factored. Is there something like the quadratic formula for cubes?
Secondly, I have the equation $3^{2x+2} − 10 \cdot 3^x + 1 = 0$. How am I supposed to get the bases to be the same. Is there some special method to use. I don't see how I can solve for $x$ without making the bases the same.
There is some sort of method for directly solving cubic equations but it's very nontrivial (http://mathworld.wolfram.com/CubicFormula.html).
That said, I think for your first question, it'll be important to note that for all $x$, $15^{-x^3+x^2-2} > 0$. This is a general property of exponential functions. With this fact in mind, how would that translate to any possible solution?
(Also a further note. I assume you took the logarithm with base $15$ of both sides? Note that the function $log_b(x)$, where $b$ is a real number, is not quite defined for $x < 0$, at least in terms of outputting real numbers. The reason why is a bit complicated to go over, though, but looking at a graph of the logarithm function should be sufficient to convince you.)
As for your second equation, notice:
$$3^{2x+2} = 3^{x+x} \cdot 3^2 = 9 \cdot 3^{x} \cdot 3^{x} = 9 \cdot (3^x)^2$$
With this rewrite completed, let $u = 3^x$ and notice that you get a quadratic in terms of $u$. Whatever the solutions are to that, you can then set them equal to $3^x$ and use logarithms to find $x$.