# Prove that the logarithmic function is continuous on R.

1. Jul 29, 2017

### Portuga

1. The problem statement, all variables and given/known data

Prove that $f\left(x\right)=\log_{a}x$ is continuous for all $\mathbb{R}$.

2. Relevant equations

I must find a $\delta>0\in\mathbb{R}$ for a given $\varepsilon>0$
such that
$$\left|x-x_{0}\right|<\delta\Rightarrow\left|\log_{a}x-\log_{a}x_{0}\right|<\varepsilon.$$

3. The attempt at a solution.

I tried to use a direct proof solving $\left|\log_{a}x-\log_{a}x_{0}\right|<\varepsilon$ for $x$. But this gives rise to a pair of values: $\delta=x_{0}\left(a^{\varepsilon}-1\right)$ and $\delta=x_{0}\left(a^{-\varepsilon}-1\right)$. When I use them to build $\left|x-x_{0}\right| < \delta$ from the inequality
$$-\delta<x-x_{0}<\delta$$
I see myself in big trouble, as there is no way to generate $\left|\log_{a}x-\log_{a}x_{0}\right| < \varepsilon$.

Can someone give me a hint, or a new strategy? I have searched for help in other forums, but hints are very sophisticated to follow.
Thanks in advance.

2. Jul 29, 2017

### Orodruin

Staff Emeritus
How do you expect a function that is not defined at $x =0$ to be continuous on the entire real line?

3. Jul 29, 2017

### Portuga

Oh, sorry, I forgot it! Continuous for all $x>0$.

4. Jul 29, 2017

### Staff: Mentor

Why don't you try and see where $\log x - \log x_0 = \log \frac{x}{x_0}$ gets you? Those proofs can often be done by writing it the wrong way and start with $|f(x)-f(x_0)| < \varepsilon$ to obtain a condition for $\delta$ to get an idea and then turn the estimations, resp. conclusions around.

5. Aug 1, 2017

### pasmith

You have solved $$-\epsilon < \log_a (x/x_0) < \epsilon$$ correctly to get $$x_0(a^{-\epsilon} - 1) < x - x_0 < x_0(a^{\epsilon} - 1),$$ but the interval is not symmetric about zero.

In this situation the closest end point to zero gives you your $\delta$. For suppose $-b < x - x_0 < c$ for strictly positive $b$ and $c$. If $b < c$ then $|x - x_0| < b$ implies $-b < x - x_0 < b < c$ as required; alternatively if $b > c$ then $|x - x_0| < c$ implies $-b < -c < x - x_0 < c$, again as required.

6. Aug 1, 2017

### Portuga

Thank you very much, pasmith! I was so close to the answer!! Now I understand. Thank you very much.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted