Show that ln(x) < sqrt(x) for x>0

[chipotleaway]

Homework Statement

Show that ln(x) < √(x) for x>0

The Attempt at a Solution

I have no idea where to start! I saw an older thread on here that asked the same question and one suggestion was to take derivatives of both functions and show that the derivative of the RHS would be larger as approaches infinity but how does this show that √(x) is greater for lower values of x?

 

[Simon Bridge]

To answer your last question - what does the derivative tell you?
If that is bigger - what does it mean?
If it were smaller - what would happen?

 

[Dick]

You also need to figure out where the derivative of log(x) starts to be greater than the derivative of sqrt(x). You are going to need a starting point for your argument.

 

[chipotleaway]

To answer your last question - what does the derivative tell you?
If that is bigger - what does it mean?
If it were smaller - what would happen?

The rate of change, so the function with the larger derivative would increase faster...I just wasn't sure using that argument alone would be enough to say that because of that, the function is larger than the other for all x>0.

I found that the derivatives are equal at x=4 and from there, it looks like the derivative of log(x) is less than the derivtaive of √(x). How would I show that this though? Do I just solve assume
1/2√(x) > 1/x and solve for x?

 

[jfgobin]

Hello ChipotleAway,

Almost there!

Now that you know that at $x=4$ the derivates are equal, you have two more steps:

• Prove that at $x=4$, $\sqrt {x} > \ln {x}$
• Prove that $\forall x > 4$, $\frac{1}{x} < \frac{1}{2\sqrt {x}}$

J.

 

[chipotleaway]

Hello ChipotleAway,

Almost there!

Now that you know that at $x=4$ the derivates are equal, you have two more steps:

• Prove that at $x=4$, $\sqrt {x} > \ln {x}$
• Prove that $\forall x > 4$, $\frac{1}{x} < \frac{1}{2\sqrt {x}}$

J.

Thanks, I'll see if I can show that! Also, wouldn't I need to somehow show that √(x) > log(x) for x between 0 and 4 as well?

 

[Mandelbroth]

Thanks, I'll see if I can show that! Also, wouldn't I need to somehow show that √(x) > log(x) for x between 0 and 4 as well?

The case where x is between 0 and 4 is considered trivial.

Both $\sqrt{x}$ and $\ln(x)$ are monotonically increasing. Because $\displaystyle \lim_{x\rightarrow 0}\ln(x)<\sqrt{0}$ and $\ln(4)<\sqrt{4}$, we know that $\ln(x)<\sqrt{x}$ from 0 to 4.

 

[mfb]

Both $\sqrt{x}$ and $\ln(x)$ are monotonically increasing. Because $\displaystyle \lim_{x\rightarrow 0}\ln(x)<\sqrt{0}$ and $\ln(4)<\sqrt{4}$, we know that $\ln(x)<\sqrt{x}$ from 0 to 4.

That is not sufficient. There are functions which satisfy all conditions you mentioned, but violate the inequality to be shown.

Instead, you can use the derivatives here as well: From 0 to 4, the derivative of ln(x) is larger than the derivative of sqrt(x)...