Where am I going wrong in solving this exponential inequality? $$(3-x)^{ 3+x }<(3-x)^{ 5x-6 }$$
Steps I took:
$$(3-x)^{ 3 }\cdot (3-x)^{ x }<(3-x)^{ -6 }\cdot (3-x)^{ 5x }$$
$$\frac { (3-x)^{ 3 }\cdot (3-x)^{ x } }{ (3-x)^{ 3 }\cdot (3-x)^{ x } } <\frac { \frac { 1 }{ (3-x)^{ 6 } } \cdot (3-x)^{ 5x } }{ (3-x)^{ 3 }\cdot (3-x)^{ x } } $$
$$1<\frac { 1 }{ (3-x)^{ 9 } } \cdot (3-x)^{ 4x }$$
$$(3-x)^{ 9 }<(3-x)^{ 4x }$$
$$4x>9$$
$$x>2.25$$
This answer seems to be wrong. I am not sure where I went wrong in the steps that I took. What did I do wrong?

Hint: use the fact that $a^x$ is decreasing  if $0<a<1$ and increasing if $a>1$.
Now consider first $2<x <3$.  In that case the inequality is satisfied iff $2<x <9/4$.  If $x>3$ there's no solution.
