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?

$$(3-x)^{ 3+x }<(3-x)^{ 5x-6 }$$
For $3>x$
$$(3+x)\log_{3-x}(3-x)<(5x-6 )\log_{3-x}(3-x)$$
$$3-x<5x-6$$
$$x>1.5$$
Thus, $$1.5<x<3$$
