1. ## Re: Differentiating Logs

Is it because elnx = xlne?

2. ## Re: Differentiating Logs

Originally Posted by Timske
Lnx=1 get rid of ln by multiplying by e

elnx = e
x=e
You would be exponentially raising each side, ie. eln(x) = e1.

3. ## Re: Differentiating Logs

Would they expect a maths adv student to do that?
Multiplying is much less complicated lol!

4. ## Re: Differentiating Logs

One last question, how would I test it? lol, whats on either side of e?

5. ## Re: Differentiating Logs

Originally Posted by D94
You would be exponentially raising each side, ie. eln(x) = e1.
Oh lol, and yes u will need to know this its not hard at all

6. ## Re: Differentiating Logs

Originally Posted by Coookies
Would they expect a maths adv student to do that?
Multiplying is much less complicated lol!
Yes, of course. What do you mean "multiplying"? There isn't multiplying :s To remove the logarithm, you take the exponents of both sides; that's the best method.

7. ## Re: Differentiating Logs

Originally Posted by Coookies
One last question, how would I test it? lol, whats on either side of e?
e is approximately 2.718..., so you can test using say 2 and 3.

8. ## Re: Differentiating Logs

So once I've raised them both, how do I get to x=e?

& I tried that but its all increasing (answer says max)

9. ## Re: Differentiating Logs

Originally Posted by Coookies
So once I've raised them both, how do I get to x=e?

& I tried that but its all increasing (answer says max)
eloge(x) = e1

Hm, it sounds like you haven't been given a good explanation yet. One of the most important concepts is to raise log by e or to raise e by log; they are sort of like inverses of each other, and can cancel each other out. But, not like multiplication/division, they are powers of each other. (You're going to need a better explanation of the whole log and e relation, because it's too easy to just take these things for granted)

Anyway, it should be a maximum; let x = 2, y' = (1 - ln(2))/4 > 0, and when x = 3, y' = (1 - ln(3))/9 < 0.

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