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.

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!

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.

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!

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.

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.

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?

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]

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!

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]