1) ln( )=2.

We know: ln( )=n ln(x)

-> 2ln(x)=ln(2) --> ln(x)= . --> x= .

Is that correct? Because my booklet, again, says something different ;)

2) are generally all logarithmic rules appliable for ln?

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- Apr 8th 2010, 06:17 AMSchderoNatural logarithms
1) ln( )=2.

We know: ln( )=n ln(x)

-> 2ln(x)=ln(2) --> ln(x)= . --> x= .

Is that correct? Because my booklet, again, says something different ;)

2) are generally all logarithmic rules appliable for ln? - Apr 8th 2010, 06:21 AMdwsmith
I obtained a this:

Multiple by 2

Raise to the e

- Apr 8th 2010, 06:22 AMlilaziz1
- Apr 8th 2010, 06:31 AMSchdero
Yes sure, sry i meant ln(

__x^2__)=2, not x^1/2. But thats no big change, because in my booklet the solution is +/-e and this is nonsense, isn't it? - Apr 8th 2010, 06:37 AMdwsmith

You can't have a negative value in the natural function since the domain is (0,infinity)

-e=-2.78......