Xjyuk

I ran into this issue during my homework. Using the rules of logarithms, I need to prove that
$$
-2lnbigg(frac2sqrt6bigg)=ln3-ln2
$$
So here were my steps:

1. First step:
   $$
   -2lnbigg(frac2sqrt6bigg)=lnbigg(frac2sqrt6bigg)^-2
   $$
   And that's as far as I got, because now I want to use the form $ln(a/b) = ln(a) - ln(b)$, but first I need to reduce the fraction because it is raised to the $-2$.

How do I evaluate $bigg(frac2sqrt6bigg)^-2$ ?

Thanks

$a^-k = frac 1a^k$ by definition.

So $bigg(frac2sqrt6bigg)^-2 =frac 1bigg(frac2sqrt6bigg)^2=$

$frac 1bigg(frac 2^2sqrt 6^2bigg)=frac sqrt 6^22^2=frac 64=frac 32$

It will help to realize that $(frac ab)^-1 = frac 1(frac ab) = frac ba$ and that $(frac ab)^k = frac a^kb^k$ to realize that that means $(frac ab)^-k = frac 1(frac ab)^k= frac 1(frac a^kb^k) = frac b^ka^k$.

(Also $(frac ab)^-k = [(frac ab)^-1]^k = (frac ba)^k=frac b^ka^k$ or that $(frac ab)^-k = frac a^-kb^-k = frac (frac 1a^k)(frac 1b^k) = frac b^ka^k$.)

In any event

$(frac 2sqrt 6)^-2 = (frac sqrt 6 2)^2 = frac sqrt 6^22^2 = frac 64 = frac 32$.

$$beginalign
-2ln left( frac2sqrt 6 right) &= -2big( ln(2)-ln(sqrt6) big) \
&= -2ln(2)+2ln(6^1/2) \
&= -2ln(2)+ln(2cdot 3) \
&=-2ln(2)+ big( ln(2)+ln(3)big) \
&=ln(3)-ln(2)
endalign$$
And for your specific question, remember that
$$left( fracab right)^-n=left( fracba right)^n$$

Using your first step,

$-2 ln(frac2sqrt6) = ln(frac2sqrt6)^-2 = ln frac1(frac2sqrt6)^2 = ln frac64 = lnfrac32 = ln 3 - ln 2$

Well you may start by distributing the index since $2$ and $sqrt 6$ are positive. Thus $$left(frac2sqrt6right)^-2=frac2^-2sqrt 6^-2.$$

Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^-n=frac 1a^n.$$ Applying this to your expression, we have $$frac2^-2sqrt 6^-2=fracfrac 12^2frac1sqrt 6^2=fracfrac 14frac16=frac64=frac 32.$$

$$ -2 ln left( frac2sqrt6 right) = ln 3 - ln 2$$

if and only if

$$ ln left( frac2sqrt6 right)^-2 = ln left( frac32right) $$

if and only if

$$ ln left[ frac1left(frac2sqrt6right)^2 right] = ln left( frac32right)$$

And so, we have

$$ ln left( frac64 right) = ln left( frac32right)$$

which is true.