Question

Can anyone explains how (1/2) + ln abs (sqrt2/2) turns into (1/2)(1-ln2)?

Answer 1

\[
\frac 12 +\ln\left|\frac{\sqrt{2}}{2}\right| = \frac 12 + \ln\left(\frac 24\right)^{1/2} = \frac 12 + \frac 12 \ln\left(\frac 12\right) = \frac 12 \left(1+\ln\left(2^{-1}\right)\right)
\]\[
=\frac 12 \left(1-\ln\left(2\right)\right)
\]

Answer 2

I dont understand how the 1/2 gets in front of ln

Answer 3

\[
\ln(a^b) = b\ln(a)
\]

Answer 4

\[
\ln(a^b) = \ln(\underbrace{a\times a\times \ldots \times a}_b) =\underbrace{ \ln(a)+\ln(a)+\ldots +\ln(a)}_b = b\ln(a)
\]

Answer 5

for some intuition, notice that
ln(x^3) = ln(x * x *x) = ln(x)+ln(x)+ln(x)= 3 ln(x)
so this property clearly holds for integer exponents. However, it is more general.. the property works for all exponents

Answer 6

It's important to note that the reason it works for real exponents is because real exponents were initially a meaningless operation, and in defining them, they were defined such that they would obey all the properties of natural number exponents. There is no coincidence.

Answer 7

Here is another way to see it: Let
\[ x= e^a \]
and
\[ \ln(x) = \ln(e^a) = a \]
then
now raise x to a power b
\[ x^b = \left(e^a\right)^b= e^{ab} \]
take the ln of both sides
\[ \ln\left(x^b\right) = ab \]
from the second line, a = ln(x) and we get
\[ \ln\left(x^b\right) = b\ \ln(x) \]