- #1

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I know I can use a graphical calculator to easily show that

How to prove ln(x) < sqrt(x) for all x>0

But I wonder if there is a rigorous way to demonstrate this.

- Thread starter grossgermany
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- #1

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I know I can use a graphical calculator to easily show that

How to prove ln(x) < sqrt(x) for all x>0

But I wonder if there is a rigorous way to demonstrate this.

- #2

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- #3

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Is there a non-graphical way of doing it?

- #4

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To answer the question, I'm assuming you have a rough idea of Calculus.

We know that the derivative of a formula, tells us the rate of change of the formula. We also that both formulas are always positive.

(Sorry no Latex I tried to make it works, but i failed. )

So let's take the derivative of ln(x) d/dx (ln(x)) = 1/x

Then of sqrt(x) d/dx (sqrt(x)) = 1/(2sqrt(x))

As x goes to infinity, 1/2 becomes useless, and for simplicity sake let's remove it.

1/x < 1/(2sqrt(x)) --> 1/x < 1/sqrt(x)

Then we can clearly see that as x increase to infinity, the rate of change of sqrt(x) is greater then ln(x).

I've just though about it while reading your question, but I am fairly sure of my answer.

Vardd

- #5

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Showing the derivative of x^1/2 – ln(x) is positive for all x>0 is fairly easy.

- #6

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You can’t just remove the ½, and your method would only show for some value of N, for x>N would ln(x) < x^1/2

To answer the question, I'm assuming you have a rough idea of Calculus.

We know that the derivative of a formula, tells us the rate of change of the formula. We also that both formulas are always positive.

(Sorry no Latex I tried to make it works, but i failed. )

So let's take the derivative of ln(x) d/dx (ln(x)) = 1/x

Then of sqrt(x) d/dx (sqrt(x)) = 1/(2sqrt(x))

As x goes to infinity, 1/2 becomes useless, and for simplicity sake let's remove it.

1/x < 1/(2sqrt(x)) --> 1/x < 1/sqrt(x)

Then we can clearly see that as x increase to infinity, the rate of change of sqrt(x) is greater then ln(x).

I've just though about it while reading your question, but I am fairly sure of my answer.

Vardd

- #7

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This is the way to do it. You also need to show that it's a positive value from the beginning though (lim x approaches 0 of f(x) = +) since increasing wouldn't really prove anything by itself (ex: any function where f(x) = - where x< some positive number k, but f(x) increasing from x>0).Showing the derivative of x^1/2 – ln(x) is positive for all x>0 is fairly easy.

- #8

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Except the statement is not trueThis is the way to do it. You also need to show that it's a positive value from the beginning though (lim x approaches 0 of f(x) = +) since increasing wouldn't really prove anything by itself (ex: any function where f(x) = - where x< some positive number k, but f(x) increasing from x>0).

0.5*1/sqrt(x)-1/x is not always positive for x>0.

- #9

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good catch, so where would the min of 0.5*1/sqrt(x)-1/x be at?Except the statement is not true

is not always positive for x>0.

- #10

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whoops, it's not monotone like I said but the minimum is positive, & that will still do it

- #11

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so that [tex]f^{\prime}[/tex] >0 for all x and therefore

[tex]e^{\sqrt{x}} - x > 0[/tex] , and thus [tex]\sqrt{x} > \ln{x}[/tex]

- #12

CRGreathouse

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whoops, it's not monotone like I said but the minimum is positive, & that will still do it

How do you prove that it's positive?

so that [tex]f^{\prime}[/tex] >0 for all x and therefore

[tex]e^{\sqrt{x}} - x > 0[/tex] , and thus [tex]\sqrt{x} > \ln{x}[/tex]

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