Question

I'm confused as to how 0 - ln(sqrt(2)/2)) = ln(2/sqrt(2)) = ln(sqrt(2))?

Answer 1

which equality is unclear?

Answer 2

The whole thing.

Answer 3

ok lets go slow

Answer 4

first of all the zero is unnecessary right?
\[0-\ln(\frac{\sqrt2}{2})=-\ln(\frac{\sqrt2}{2})\]

Answer 5

now is it clear that \[-\ln(x)=\ln(\frac{1}{x})\]?

Answer 6

yes.

Answer 7

there are lots of explanations if it is not obvious to you
one is that \[\ln(x^n)=n\ln(x)\] so \[-1\ln(x)=\ln(x^{-1})=\ln(\frac{1}{x})\] but there are other ways to see it

Answer 8

ok so if \(x=\frac{\sqrt2}{2}\) then \[\frac{1}{x}=\frac{2}{\sqrt2}\]that should make the first equality clear

Answer 9

Yes. I understand to that part now.

Answer 10

ok then for the next equality, it is always true that \[\frac{a}{\sqrt a}=\sqrt{a}\]

Answer 11

there are lots of ways to see that too
one is two square \[\frac{a}{\sqrt{a}}\] and see that you get \(a\)

Answer 12

that last part is called rationalizing the denominator.

Answer 13

another is to use exponents \[\large \frac{a^1}{a^{\frac{1}{2}}}=a^{1-\frac{1}{2}}=a^{\frac{1}{2}}\]

Answer 14

I see

Answer 15

and yet another way, as @mww said is to multiply top and bottom by \(\sqrt{a}\) aka rationalize the denominator

Answer 16

I always get stuck with algebra. Thanks!

Answer 17

that is everyone's problem...
yw