Answer
You can only apply the uniqueness of $b^x$ to solve an exponential equation if the same base can be determined.
Work Step by Step
As an example, we can use the uniqueness of $b^x$ to solve the following equation: $4^{x+3}=32^{x-1}$
$4^{x+3}=32^{x-1}$
$2^{2*(x+3)}=2^{5*(x-1)}$
$2^{2x+6}=2^{5x-5}$
Using the uniqueness property, we have the following:
$2x+6=5x-5$
$2x+6-2x+5=5x-5-2x+5$
$11=3x$
$11/3=3x/3$
$11/3=x$
As an another example, we cannot use the uniqueness of $b^x$ to solve the following equation: $3^{x+4} = 6^{x-1}$.
We cannot use the uniqueness since $6=2*3$, and the other side of the equation doesn't have a factor of $2$.
