I need to prove the continuity of $f(x)=\log x$ using a $\epsilon-\delta$ proof

By your inequality, the absolute value of the difference is $\lt \epsilon$ if $$\frac{a}{e^{\epsilon}}-a \lt x-a\lt ae^\epsilon -a$$ (we subtracted $a$ from each side of each of your two inequalities). Let $\delta=a\min\left(1-\frac{1}{e^{\epsilon}}, e^\epsilon -1\right)$.

Remark: Actually, $1-\frac{1}{e^{\epsilon}}$ is the smaller of the two, so in effect we are letting that be $\delta$. But we really don't need to bother finding that out: all we need to do is to show there is a $\delta$ that works.

Hint: from the right (i.e., $\,x>a\,$):

$$|\log x-\log a|=\log\frac{x}{a}<\epsilon\Longleftrightarrow \frac{x}{a}<e^\epsilon\Longleftrightarrow x<ae^\epsilon\;\;(\text{remember}:\;a,x>0\,\;!)\Longrightarrow$$

$$x-a<a(e^\epsilon -1)$$

and there you have your $\,\delta>0\ldots\,$