Is this logarithm-based proof rigorous? Since I'm not very familiar with logarithms I'm not sure if this proof is correct. The problem is this:
Prove that if $a,b > 0$ and $a>b$ then is true that $a^a>b^a$ (this comes from another post)
My proof is this:
We can take the $\log$ in base $a$ of both $a^a$ and $b^a$ since they're both positive quantities
$$\log_a(a^a)>\log_a(b^a)$$
$$a>\log_a(b^a)$$
Now we can change the base of $\log_a(b^a)$ ti $b$ using the change-of-base formula
$$\log_a(b^a)=\frac{\log_b(b^a)}{\log_b(a)}=\frac{a}{\log_b(a)}$$
If we substitute in the last equation we get
$$a>\frac{a}{\log_b(a)}$$
$$\log_b(a)>1$$
Which is always true because $b<a$ QED
Is this correct?
 A: In principle, this is fine. 
One remark: Make sure (and express) that all steps are equivalence transforms. 
Another remark:  Note that it does not matter to which base ($a$ or $b$ or anything) you take the log. Can you see that?
Final remark: It may be simpler to expand $\log(b^a)=a\log b$ earlier.
A: It seems correct but what you have written is the path to the discovery of a proof, not the proof itself. You need to write your steps in reverse order to make it a proof, taking care to justify all steps.
A: I'm going to disagree with the other answers. It is not correct because you manipulate inequalities in a way which assumes $a,b>1$, whereas you are supposed to be showing the result for $a>b>0$. 
For example, from $a^a>b^a$ you go to $\log_a(a^a)>\log_a(b^a)$. This assumes $a>1$ - try $a=1/2$, $b=1/4$ and then $a^a>b^a$ but $\log_a(a^a)=1/2<1=\log_a(b^a)$.
To do it this way you need to consider the cases $1<b<a$ and $0<b<1<a$ and $0<b<a<1$ (not to mention cases where $a=1$ or $b=1$, since $\log_1x$ is meaningless) separately. (The first case is fine as it is; in the second and third there are two places where the inequality swaps round - not the same two places! - so the errors cancel out.) Or, as Hagen's answer suggests, you could use some fixed base $>1$ which doesn't depend on $a$ and $b$ to avoid the issue.
