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 April 12th, 2018, 02:16 PM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Why is logx monotonic Definition: f(x) is strictly increasing if f(x) increases iff x increases. Then if f(x) is strictly increasing f(f$\displaystyle ^{-1}$(x)) = x implies f$\displaystyle ^{-1}$(x) is strictly increasing. Example: e$\displaystyle ^{logx}$=x
 April 12th, 2018, 02:29 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 I think it would be simpler to note that, since y= ln(x) is the inverse of [math]x= e^y[\math], [math]\frac{dy}{dx}= \frac{1}{\frac{dx}{dy}= \frac{1}{e^x}[\math]. Since [math]e^y[math] is non-decreasing, its derivative is non-negative so that [math]\frac{d ln(x)}{dx}[\math] is non-negative so non-decreasing.
April 12th, 2018, 03:18 PM   #3
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 Originally Posted by Country Boy I think it would be simpler to note that, since y= ln(x) is the inverse of [math]x= e^y[\math], [math]\frac{dy}{dx}= \frac{1}{\frac{dx}{dy}= \frac{1}{e^x}[\math]. Since [math]e^y[math] is non-decreasing, its derivative is non-negative so that [math]\frac{d ln(x)}{dx}[\math] is non-negative so non-decreasing.
That is way more complicated then it needs to be. You already need quite some structure, like the derivatives of log(x) and e^x, the inverse function theorem, etc. These are already some facts that might be tricky to establish!

But in your proof you already use that e^x is nondecreasing. Doesn't this imply directly that the "inverse" log(x) is nondecreasing without needing all the machinery of derivatives?

In any case, the OP is raising an interesting question. But sadly it depends quite crucially on how log(x) is defined to begin with. The traditional way in high schools is to define it as the "inverse" of e^x, in which case I think the "proof" in the OP is optimal. But historically, and in many math books a different approach is taken. For example, in a lot of book they will define
$$\log(x) =\int_1^x \frac{1}{t}dt$$
(and then they will actually use this to define e^x, because rigorously defining e^x can actually be somewhat tricky to do it directly).
If log(x) is defined as an integral in this way, then Country boy's use of the positivity of the derivative is good, although a direct proof not involving derivatives is definitely also possible.

 April 12th, 2018, 03:45 PM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 Of course the first thing you think of is positive derivative. But that involves another concept, and I asked myself why does that imply monotonicity, aside from intuition. I vaguely recall a proof using mean value theorem, which rules it out, because I couldn't recall it. And there is the jagged line. Mirror image of the function is graphical and not immediate, why is the mirror image monotonic. You have to think about what it means graphically. If the OP doesn't feel right, that's a perfectly valid objection, which I appreciate. What is simple, logical and worth remembering (easily) to me might be blah blah to someone else. Just happened to wonder why logx was monotonic, and the first (second) thing I did was write down the definiition, e$\displaystyle ^{logx}$ =x and then the OP ocurred to me.
 April 13th, 2018, 12:16 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 I'm really going beyond log x, and I wanted to avoid details of definition. If f(x) is monotonic and has an inverse, then f(f$\displaystyle ^{-1}$(x)) = x, which says: If x increases argument of f increases, and if argument of f increases, x increases. The argument of f is f$\displaystyle ^{-1}$(x). EDIT With reference to cjem's post, if according to your definition logx is monotonic and has an inverse, f(f$\displaystyle ^{-1}$(x))=log(e$\displaystyle ^{x}$)=x and e$\displaystyle ^{x}$ is monotonic. Last edited by zylo; April 13th, 2018 at 12:31 PM.

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