Differentiating Logs

[Coookies]

Is it because elnx = xlne?

[D94]

Lnx=1 get rid of ln by multiplying by e

elnx = e
x=e

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

[Coookies]

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

[Coookies]

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

[Timske]

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

Oh lol, and yes u will need to know this its not hard at all

[D94]

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.

[D94]

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.

[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)

[D94]

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

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

e^log_e(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.