Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$ 
Prove that if $2b^2+1 = 3^b$ where $b$ is a positive integer, then $b \leq 2$.

Is there some way we can transform the equation in order to get the inequality? We have $2b^2 = 3^b-1$.

Subtract $b^3$ from both sides.
$2b^2 + 1 -b^3 = 3^b-b^3$
$\Rightarrow b^2(2-b) = 3^b - (b^3+1)$
Now think about the expression 
$3^b - (b^3+1)$. 
Both $3^b$ and $b^3+1$ are monotonically increasing. Being an exponential function, $3^b$ grows faster than $b^3+1$ for positive b. These two facts ensure that if at any non-negative point $3^b$ dominates $b^3+1$, it will continue to dominate $b^3+1$ at subsequent points. And as a matter of fact, $3^b=b^3+1$ for $b=0$ 
Thus, for $b\geqslant0$ we have $3^b-(b^3+1)\geqslant0$
$\Rightarrow b^2(2-b)\geqslant0$
$\Rightarrow 2 \geqslant b$
