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  • 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

Community
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onlyme
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    What definition of the logarithm function do you have? – Jack D'Aurizio Feb 04 '15 at 18:43
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    That is not a definition. This Cauchy functional equation has infinitely many solutions, most of them decidedly non-continuous. And even the continuous ones are not unique, every logarithm $\log_a$ is a solution. Thus your solution heavily depends on the definition of the natural logarithm or alternatively of THE exponential function. – Lutz Lehmann Feb 04 '15 at 19:53
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    Really ? That's awesome. I didn't know this. Do you know a good way to define the logarithm ? To convert the base is just to divide by a constant so it's not that hard to prove that every logarithm is continuous if you prove that one of them is. – onlyme Feb 04 '15 at 19:59
  • The question is not what a good definition is, but what the definition is that your course uses and that your corrector will base the evaluation of your homework on. Typically modern treatments define exponential functions via integer, then rational powers and then Dedekind cuts and define the logarithm as inverse functions of the exponential functions. Then the Euler number and natural logarithm happen as quasi-accident. Later the characterization as "natural" is justified via power series. – Lutz Lehmann Feb 04 '15 at 21:22
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    I'm not even in a college right now so don't need worry with "my course definition". Thought you pointed me that most courses don't have a good definition of these functions. That limit definition is quite cool I would like to see some proofs on that. – onlyme Feb 04 '15 at 21:26
  • Set $h=1/n$ and you get the difference quotient for the exponential function (which, as well as l'Hopital, might be circular in this case). Set $x=1+z$ and use the binomial series to get the power series of the logarithm in the limit of the expansion (one has to justify why the two limits commute). – Lutz Lehmann Feb 06 '15 at 19:16
  • And I'm quite sure that most courses have good. even if disjointed, definitions of these functions. If the progression from rational powers $a^{m/n}$, to exponential functions $a^x$, the natural basis or Euler number $e=\lim (1+1/n)^n$ and finally to the equality of $e^x$ and $\exp(x)=\sum x^k/k!$ is as well received by the students is a different question. With that firmly established one can also use the strong monotonicity of $e^x$ to prove the continuity of its inverse function $\ln(x)$. – Lutz Lehmann Feb 06 '15 at 19:24
  • See the full development of the theory of logarithmic function based on the limit definition at http://paramanands.blogspot.com/2014/05/theories-of-exponential-and-logarithmic-functions-part-2_10.html – Paramanand Singh Jan 14 '17 at 06:04

2 Answers2

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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.

Jack D'Aurizio
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  • How do you know that the natural logarithm is the solution of this functional equation? At what point are the cloud solutions excluded? – Lutz Lehmann Feb 04 '15 at 19:54
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    @LutzL: we know that the logarithm satisfies such functional equations since its inverse function (the exponential function) satisfies $$ f(x)f(y)=f(x+y).$$ What do you mean by "cloud solutions"? – Jack D'Aurizio Feb 04 '15 at 20:24
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    There are two ways to read the answer. Starting with the logarithm as the inverse function of the exponential, all is clear and correct, assuming the (as of now unspecified) definition of the exponential function easily gives the Bernoulli inequality. In the question however it seems that the functional equation is the point of departure, and without some continuity assumption, the common solution of that is everywhere dense in the upper half-plane, thus "cloudy". – Lutz Lehmann Feb 04 '15 at 21:15
  • I am not using the functional equation with some hidden assumption like continuity. I am just stating that the functional equation for the logarithm comes from the functional equation for its inverse function, and that the functional equation provides a way for proving continuity over $\mathbb{R}^+$ by just proving continuity in a point. Continuity in $x=1$ follows from standard inequalities. – Jack D'Aurizio Feb 04 '15 at 21:18
  • I'm sorry if it was unclear, I was referring to what known facts and definitions the OP has at their disposition, especially regarding the "naturalness" of the natural logarithm and Euler number. In the common context, there is nothing to critizise. – Lutz Lehmann Feb 04 '15 at 21:26
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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.

Olod
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