Which of these Two Functions is Greater?

[Luminal]

Which of these two tend toward zero slower? In other words, which one will eventually remain larger than the other for all x greater than an arbitrary finite number?

 

f(x) = 1/(x(lnx)^2)

 

f(x) = 1/((ln(3^x))^2)

 

Sorry for the prolific use of parentheses; I wanted to make the functions very clear.

[thedarkshade]

I can't really figure out how's the real form of the functions but I'll try.

if the first one is:

[math]f(x)=\frac{1}{xlogx^2}[/math]

then this can also be written as:

[math]f(x)=-2xlogx[/math]

 

and if the second function is:

[math]f(x)=\frac{1}{(log3^x)^2}[/math]

then this can also be written as:

[math]f(x)=-2xlog3[/math]

 

so now you got:

1. [math]f(x)=-2xlogx[/math] and

2. [math]f(x)=-2xlog3[/math]

 

now give x any value and you'll find out what you want!

[Cap'n Refsmmat]

I'm not quite sure what you just did there. You seem to have jumped from

[math]f(x) = \frac{1}{x (\ln x)^2}[/math]

to

[math]f(x) = \frac{1}{2x \ln x}[/math]

implicitly, which doesn't work because we're taking the ln of x and squaring it, not the ln of x squared.

[thedarkshade]

I'm not quite sure what you just did there. You seem to have jumped from

[math]f(x) = \frac{1}{x (\ln x)^2}[/math]

to

[math]f(x) = \frac{1}{2x \ln x}[/math]

implicitly, which doesn't work because we're taking the ln of x and squaring it, not the ln of x squared.

 

No no,, it's different!

[math]f(x) = \frac{1}{x (\ln x)^2}[/math]

you just switch sides of the fraction and you get:

[math]f(x)={x (\ln x)^2}^{-1}[/math]

now that -1 multiplies that square, so no we get:

[math]f(x)=x (\ln x)^{-2}[/math]

then -2 goes in front and we get

[math]f(x)=-2xlnx[/math]

I also did this on the second function too!

[Cap'n Refsmmat]

The exponent can only go in front if it's part of the natural logarithm.

[math]x^4 \neq 4x[/math]

 

The exponent in that case is outside of the parentheses and you can't split it out, though I might be wrong.

[Luminal]

Anyway, this determines if I get an A or B in Calculus II.

 

I claimed f(x) = 1/((ln(3^x))^2) would always remain larger (or alternatively, the bottom smaller) than f(x) = 1/(x(lnx)^2) after 5 (arbitary finite number) to prove a Series converged, by Direct Comparison of course. Long story short, he said I was wrong and I do not know if he came about accurately at his conclusion.

 

Oh, and I only have a day or so remaining to correct him before my grade is final. Expedience would be greatly valued.

[Mr Skeptic]

Which of these two tend toward zero slower? In other words, which one will eventually remain larger than the other for all x greater than an arbitrary finite number?

 

f(x) = 1/(x(lnx)^2)

 

f(x) = 1/((ln(3^x))^2)

 

Sorry for the prolific use of parentheses; I wanted to make the functions very clear.

 

Did you mean [math]f(x) = \frac{1}{x (ln x)^2}[/math] and [math]f(x) = \frac{1}{(ln (3^x))^2}[/math]? If you quote someone's post, you can get a copy of the math language used, so you don't need so many parenthesis.

 

Now [math]\frac{1}{(ln (3^x))^2} = \frac{1}{(x ln 3)^2} = \frac{1}{x^2 (ln 3)^2}[/math], so this decreases as [math]\frac{1}{x^2}[/math], whereas the other decreases by [math]\frac{1}{x (ln x)^2}[/math]. So the question becomes which increases faster, x or (ln x)^2?

[thedarkshade]

The exponent can only go in front if it's part of the natural logarithm.

[math]x^4 \neq 4x[/math]

 

The exponent in that case is outside of the parentheses and you can't split it out, though I might be wrong.

here's the rule:

[math]loga^k=k\times loga[/math]

[DarkNebula]

here's the rule:

[math]loga^k=k\times loga[/math]

 

Yeah but that only works for things inside the log function.

[math](log(n))^2 [/math] (or [math]log(n) * log(n)[/math] ) is not the same as [math]log(n^2)[/math] at all.

 

[math]log(3^2) = 2 log(3)[/math] is true

but what you were doing just doesn't work. Try it for yourself,

plug a number in there even!

[Bignose]

here's the rule:

[math]loga^k=k\times loga[/math]

 

Sure, but you have to be more careful with the parentheses:

 

[math]\log{a^k} \neq (\log{a})^k[/math]

 

Like was said above [math](\log{a})^2 = (\log{a})*(\log(a))[/math] not [math]2*\log{a}[/math]