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Say we wanted to prove the continuity of the logarithm using the $\delta - \epsilon$ proof (and using the definition of the log as the inverse of the exponential). For any log base $a>1$ Starting with $|\log_a({x_1}) - \log_a({x_2})|<\epsilon$, We can find that if $\frac{x_1}{x_2} < a^\epsilon$ (and $\frac{x_1}{x_2}<a$), then $|\log_a({x_1}) - \log_a({x_2})|<\epsilon$ as desired. But for the $\delta$ part to come in, we need it in the form of $|x_1 - x_2|$, not $\frac{x_1}{x_2}$.

So then I thought that maybe if $|x_1 - x_2| < \frac{x_1}{x_2} -1 < a^\epsilon -1$ then we would have $|\log_a({x_1}) - \log_a({x_2})|<\epsilon$ which would mean that the function is continuous at $x_1$ and since it was arbitrary it is continuous at all x (right?).

The reason I think this is true is because we want the distance between $x_1$ and $x_2$ to be very small, so $a^\epsilon -1$ (which we would call $\delta$) would get really small as $\epsilon$ gets small. Is there a better way to arrive at this value of $\delta$? Assuming it is a valid answer

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    $\begingroup$ What definition of $\log$ are you using? One way this can be done is to define $\log$ as $\log(x)=\int_1^x\frac1t\mathrm{d}t$, and then use continuity of Riemann integrals. Generally the continuity of the logarithmic function is not proved with a standard $\delta/\epsilon$ proof $\endgroup$ – Václav Mordvinov Oct 11 '18 at 20:10
  • $\begingroup$ We are using the inverse of the exponential definition. We haven't covered any other methods in my class so far so I'm hoping to answer it with $\delta - \epsilon$ $\endgroup$ – Riley H Oct 11 '18 at 20:15
  • $\begingroup$ @RileyH How are you defining the exponential function? $\endgroup$ – Jack M Oct 11 '18 at 20:16
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What you propose is the following proof. In what follows I'll assume $a>1$, I can't be bothered to worry about whether or not the details apply if $a\leq 1$.

Let $\epsilon>0$ and $x_0>0$. We want to find a small enough $\delta$ such that for $|x-x_0|<\delta$, $|\log_a(x)-\log_a(x_0)|<\epsilon$. We have

$$|\log_a(x)-\log_a(x_0)|=|\log_a(\frac x{x_0})|$$

and you suggest that if we can guarantee $\frac x{x_0}<a^\epsilon$, then we're done. First of all, this is completely untrue as stated because that could mean $\frac x{x_0}$ could be close to $0$ in which case that logarithm would be massively negative, not what we want. What we want is for $\frac x{x_0}$ to be close to $1$. A better condition would be $a^{-\epsilon}<\frac x{x_0}<a^\epsilon$. In that case, we would have, by the strictly increasing nature of the $\log$ function:

$$-\epsilon<\log_a(\frac x{x_0})<\epsilon$$ $$|\log_a(\frac x{x_0})|<|\epsilon|$$

As you say, we need to show that we can guarantee $a^{-\epsilon}<\frac x{x_0}<a^\epsilon$ for $|x-x_0|<\delta$ by choosing a small enough $\delta$. Probably the cleanest way to do this is to separate the cases $x>x_0$ and $x<x_0$. For $x>x_0$ you just need $\frac x{x_0}<a^\epsilon$ since the other inequality is trivial, letting $\delta=a^\epsilon-1$ we get immediately $|\frac x{x_0}-1|<\delta$, the case $x<x_0$ is similar. So that part of the proof isn't that mysterious.

The mysterious part is figuring out if this is really a proof. What facts about logarithms and exponentials did we implicitly use here? Can we justify all of them? We used the following:

  1. $\log_a$ is a real function defined for all $x>0$.
  2. The formula $\log(x)-\log(y)=\log\frac{x}{y}$.
  3. $\log$ is strictly increasing.
  4. $a^x$ is strictly increasing (this is equivalent to (3) under your definitions).
  5. $a^0=1$.

I was originally going to write an answer about how we would first have to prove that $a^x$ is continuous, but I don't think I've used that assumption anywhere. I suppose the key is in assumption (2), which is such a powerful assumption (it basically uniquely characterizes the logarithm) that it implicitly gives us the continuity of $\log$ automatically. In fact, I think it should be possible to skip any mention of $a^x$ and show $\log$ is continuous just using the assumption $\log(x)+\log(y)=\log(xy)$.

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