Question

Use the logarithmic properties to expand ln(x^2-16/x^4), where x>4

Answer 1

Are you SURE you don't mean \(\dfrac{x^{2} - 16}{x^{4}}\)? Just checking.

Answer 2

Yes, with parenthesis.

Answer 3

I'm not certain you clarified it. Which is it?

\(\ln\left(x^{2} - \dfrac{16}{x^{4}}\right)\) as you have written it, or otherwise, perhaps \(\ln\left(\dfrac{x^{2} - 16}{x^{4}}\right)\)

Answer 4

the second one you wrote.

Answer 5

Do you see why that is NOT what you have written? It is important that you understand this. It is a matter of the Order of Operations.

Factor the numerator and see if anything dawns on you.

Answer 6

Sorry, it was hard to write it what I wanted.

Answer 7

It only requires additional parentheses. I'm guessing you DO know how to do that. Take the time to write clearly.

Now, how does that numerator factor?

Answer 8

factor the numerator and the denominator

Answer 9

Well, yes, but factoring the denominator is trivial.

Answer 10

HINT: to factor the numerator observe that it is in fact a difference of squares

Answer 11

And, just for the record, factoring the Difference of Squares should be ALMOST trivial by the time you get to the calculus. So, factor away. Let's see what you get.

Answer 12

and what about x>4?

Answer 13

We can talk about that later. Factor!

Answer 14

Okay.

Answer 15

(x^4)(x^2-16)=x^6-16x^4?

Answer 16

\(\dfrac{x^{2} - 16}{x^{4}} = \dfrac{(x+4)(x-4)}{x^{4}}\)

What were you doing? How did that \(x^{4}\) get into the numerator?

Answer 17

I'm sorry /.\ my teacher haven't teach me yet.

Answer 18

she gave me during winter break.

Answer 19

Not sure where to go with that. If you cannot factor a difference of squares, you cannot solve this problem. You'll have to wait until you cover that material.

From the logarithm point of view, we have this well-known rule:

log(a*b) = log(a) + log(b)

This is not ALWAYS the case unless you KNOW that a and b are positive. This completely explains the restriction x>4.

Answer 20

I know the log(a*b)=log(a)+log(b)

Answer 21

That's nice, but that rule is of no value whatsoever on this problem unless you can factor the numerator. Kinda' stuck where we are.

Answer 22

aw, shoot. :/ Well, thanks for helping me and I am sorry I am wasting your time.

Answer 23

Nope, no time wasting at all. It is good to figure out where you are. Now we know and you know where you need a little more local help. We can't fix everything, here.

Hang in there. You'll get it. :-)

Answer 24

Thank you so much. :) Have a nice day.