Question

Why is ln(1/2) = -ln(2)

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset1prb.pdf
1H-1A

Solution:
http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-b-implicit-differentiation-and-inverse-functions/problem-set-2/MIT18_01SC_pset1sol.pdf
1H-1A

Answer 1

because (for any base not only ln) log\[\log_{a} ^{x^b} = b \log_{a} x\]
the power b comes in front of logarithm

Answer 2

Ahh I finally got it.
\[\ln(1/2)=\ln(1)-\ln(2)=-\ln(2)\]
since ln(1) = 0.
Or is yours a different approach?

Answer 3

\[\ln{\frac{1}{2} = \ln({1})-\ln({2}); \ln({1})=0; -\ln(2) =\ln({2^{-1})}}; 2^{-1}= \frac{1}{2}\]