how to solve this equation using logarithms for the equation :
$ 3^{x^2} + 3^x = 90 $
 my solution was :
$3^{x^2} + 3  ^x = 3^{2^2} + 3^2$
so $x=2$
but i want to know if there is any solution by using logarithms ?
when using wolframAlpha th solution was $ x= -2.02356 $
or $x = 2 $
but How ? 

If we divide both sides by $3$ first, we get $$3^{x^2-1} + 3^{x-1} = 30$$
Then, we can write it as $$(3^{x-1})^{x+1}+3^{x-1} = 3^{x-1}(3^{x-1})^{x}+3^{x-1}= 30$$
Now, if we bracket $3^{x-1}$, we get $$3^{x-1}(3^{x^2-x}+1)=30$$
From here, if there exists an integer solution, $30$ must be factorized as $3 \cdot 10$ so that $3^{x-1}$ is an integer (Notice that in other factorizations of $30$, there is no term $3^k$ where $k \in \mathbb{Z^+}$). So we can try $3^{x-1} = 3 \implies x=2$ and we see that it is a solution of this equation since $3^{x^2-x}+1 = 10$ when $x=2$.
