# Natural logarithms

• Apr 8th 2010, 06:17 AM
Schdero
Natural logarithms
1) ln($\displaystyle x^{\frac{1}{2}}$)=2.
We know: ln($\displaystyle x^{n}$)=n ln(x)
-> 2ln(x)=ln(2) --> ln(x)=$\displaystyle {\frac{ln(2)}{2}}$. --> x= $\displaystyle e^{\frac{ln(2)}{2}}$.
Is that correct? Because my booklet, again, says something different ;)

2) are generally all logarithmic rules appliable for ln?
• Apr 8th 2010, 06:21 AM
dwsmith
I obtained a this:
$\displaystyle \frac {1}{2}*ln(x)=2$

Multiple by 2

$\displaystyle ln(x)=4$

Raise to the e

$\displaystyle x={e}^{4}$
• Apr 8th 2010, 06:22 AM
lilaziz1
Quote:

Originally Posted by Schdero
1) ln($\displaystyle x^{\frac{1}{2}}$)=2.
We know: ln($\displaystyle x^{n}$)=n ln(x)
-> 2ln(x)=ln(2) --> ln(x)=$\displaystyle {\frac{ln(2)}{2}}$. --> x= $\displaystyle e^{\frac{ln(2)}{2}}$.
Is that correct? Because my booklet, again, says something different ;)

2) are generally all logarithmic rules appliable for ln?

2. yes because ln(x) is the same thing as $\displaystyle \log_e(x)$
• Apr 8th 2010, 06:31 AM
Schdero
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 AM
dwsmith
$\displaystyle ln(x^{2})=2$

$\displaystyle 2ln(x)=2$

$\displaystyle ln(x)=1$

$\displaystyle x=e$

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

-e=-2.78......