Prove that if $x>1$, then $log_a(x)>0$

Question

If $x>1$ then $log_a(x)>0$

Well I thought that log with base a of 1 is 0. I don't know what to do more.

Answer 1

$$\log_a(x)=\frac{\ln(x)}{\ln(a)}$$

Now, you have $\log_a(x)\gt 0$ iff you have one of the following two cases:

$$ i)~\ln(x)\gt 0~\land~\ln(a)\gt 0\\ ii)~\ln(x)\lt 0~\land~\ln(a)\lt 0$$

Since $x\gt 1$, you already have $\ln(x)\gt 0$. So, we need to get case (i).

For that, we need to have $\ln(a)\gt 0\implies a\in(1,+\infty)$

If it happens, the result is trivial. Otherwise, the claim in your question is false.

Answer 2

In general:

$$\log_ax=y>0\iff\begin{cases}a>1\;\;and\;\;x>1\\{}\\0<a<1\;\;and\;\;0<x<1\end{cases}$$