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?
 A: 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.
A: The first condition you have to assume is that $3-x>0$, for the expressions to be meaningful.
Now, take the natural logarithm of both sides; since the natural logarithm is an increasing function, the inequality is preserved:
$$
(3+x)\log(3-x)<(5x-6)\log(3-x)
$$
One could now be tempted to remove the $\log(3-x)$ term from both sides, but it would be wrong, because multiplying by a negative number reverses inequalities. So we have to make three cases:
Case 1: $\log(3-x)<0$
This means $3-x<1$ (that is, $x>2$) and the inequality becomes
$$
3+x>5x-6
$$
or $4x<9$. This gives a solution set
$$
2<x<\frac{9}{4}
$$
Case 2: $\log(3-x)=0$
There is no solution for $x=4$, because the inequality would be $0<0$, that's false.
Case 3: $\log(3-x)>0$
This means $3-x>1$ (that is, $x<2$). The inequality becomes
$$
3+x<5x-6
$$
or $4x>9$. The two conditions are incompatible, no solution here.
Final step
We must put together the found solution sets (just one, in this particular case) with the top condition. This provides the final solution
$$
\boxed{2<x<\frac{9}{4}}
$$
Note
It's actually immaterial what base the logarithm is, as long as it's greater than $1$, so the logarithm is increasing. An error would be using $\log_{(3-x)}$, because it's unknown whether it's increasing or decreasing or even existing, unless one makes a division into cases as before. Using a fixed basis is easier, in my opinion.
Where did you go wrong? In not considering that $3-x$ may be positive or negative and also in not considering that the exponential function is increasing when the basis is greater than $1$, but decreasing when the basis is in the interval $(0,1)$, while it is constant for the basis $1$ (and undefined for a negative basis).
A: $$(3-x)^{ 3+x }<(3-x)^{ 5x-6 }$$
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, $$1.5<x<3$$
