How do I prove using the definition that the logarithmic function is continuous 
*

*Do not define the logarithmic function as an integral.

*Do not use that $e^x>x$ because this assumes the continuity of the exponential. 


I have no clue on how to solve this one.
I guess I got it, but i'll leave this question here in case someone else has the same problem.
$|\ln x - \ln a|<\epsilon \rightarrow -\epsilon < \ln \frac x a < \epsilon$
$ae^{-\epsilon} - a < x-a < ae^{\epsilon} - a$
(of course $a>0$) taking
$\delta = \min \{ a(1 - e^{-\epsilon}), a(e^{\epsilon} -1) \} = a(1 - e^{-\epsilon})$
we're done. I guess. If anybody has another way of proofing please show me.
The definitions of the logarithm function can be these:
$\log : (0, +\infty) \rightarrow \mathbb{R} $
$ \log(x) + \log(y) = \log(xy) , \forall (x,y)$ both real greater then zero.
Of course some basic properties come from this definition and you can use them.
But you can also define it as the inverse of the exponential as long as you don't use the continuity of the inverse to prove it.
A better definition may be:
$\ln(x) = \lim_{n \rightarrow \infty} n(x^{\frac 1 n} -1)$
I need to prove the continuity of $f(x)=\log x$ using a $\epsilon-\delta$ proof
 A: Hint: Since the logarithm function satisfies
$$ f(xy) = f(x)+f(y) $$
for any $x,y\in\mathbb{R}^+$, in order to prove the continuity over $\mathbb{R}^+$ you just need to prove the continuity in $1$, since:
$$ \log(x+\varepsilon)-\log(x) = \log\left(1+\frac{\varepsilon}{x}\right).$$
Now the continuity in $1$ follows from the Bernoulli inequality:
$$ \forall x\in(-1,1),\quad x+1\leq e^x \leq \frac{x}{x-1} \tag{1}$$
(proving $(1)$ does not necessarily depends on the continuity of the exponential function. For instance, we can prove $(1)$ for any $x\in(-1,1)\cap\mathbb{Q}$ by induction) and a straightfoward consequence of $(1)$ is:
$$ \frac{y}{1+y}\leq \log(1+y) \leq y \tag{2} $$
for any $y$ in a neighbourhood of zero. Continuity follows.
A: Suppose that $a > 1.$ We wish to prove that the logarithmic function
$$
f(x)=\log_a(x)
$$
is continuous at $1.$ 
Let $\varepsilon > 0$ be any positive number. Take a natural number $n$ such that 
$$
\frac 1n < \varepsilon 
$$
and set
$$
\delta_1 = \sqrt[n]a-1 > 0. 
$$
Then 
$$
1 < x < 1+\delta_1 =\sqrt[n]a
$$
implies that
$$
0 < \log_a(x) < \frac 1n < \varepsilon. 
$$
In effect, 
$$
\lim_{x \to 1_+} \log_a(x)=0.
$$
Similarly, set
$$
\delta_2=1-\sqrt[n]{\frac 1a} > 0.
$$
Then
$$
\sqrt[n]{\frac 1a}=1-\delta_2 < x < 1
$$
implies that 
$$
-\varepsilon < -\frac 1n < \log_a(x) < 0,
$$
whence
$$
\lim_{x \to 1_-} \log_a(x)=0.
$$
Consequently, as the one-sided limits as $x \to 1$ exist and equal to each other,
$$
\lim_{x \to 1} \log_a(x)=\log_a(1)=0,
$$
and the logarithmic function
$$
f(x)=\log_a(x)
$$
is continuous at $1.$ 
As it has been noted above, continuity of $f(x)$ at $1$ implies continuity at all points of the domain of $f.$ Finally, since
$$
\log_{1/a}(x)=-\log_a(x) \qquad (x > 0)
$$
any logarithmic function whose (positive) base is less than $1$ is also continuous. 
