Is this logarithm-based proof rigorous?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.