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I ran into this issue during my homework. Using the rules of logarithms, I need to prove that $$ -2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln3-\ln2 $$ So here were my steps:

  1. First step: $$ -2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln\left(\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}\right) $$ 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 $\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}$ ?

Thanks

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    $\begingroup$ Note that $x^{-a} = \frac{1}{x^a}$ $\endgroup$ – desiigner Jul 11 at 19:53
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By definition $$a^{-k} = \frac 1{a^k}$$

So $$\left(\frac{2}{\sqrt{6}}\right)^{-2} =\frac 1{\left(\frac{2}{\sqrt{6}}\right)^{2}}=$$

$$\frac 1{\left(\frac {2^2}{\sqrt 6^2}\right)}=\frac {\sqrt 6^2}{2^2}=\frac 64=\frac 32$$

It will help to realize that $(\frac ab)^{-1} = 1/(a/b) = \frac ba$ and that $(\frac ab)^k = \frac {a^k}{b^k}$ to realize that that means $$\left(\frac ab\right)^{-k} = \frac 1{\left(\frac ab\right)^k}= \frac 1{\left(\frac {a^k}{b^k}\right)} = \frac {b^k}{a^k}.$$

(Also $(\frac ab)^{-k} = [(\frac ab)^{-1}]^k = (\frac ba)^k=\frac {b^k}{a^k}$ or that $(\frac ab)^{-k} = \frac {a^{-k}}{b^{-k}} = (1/a^k)/(1/b^k) = \frac {b^k}{a^k}$.)

In any event

$$\left(\frac {2}{\sqrt 6}\right)^{-2} = \left(\frac {\sqrt 6}{ 2}\right)^2 = \frac {\sqrt 6^2}{2^2} = \frac 64 = \frac 32.$$

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    $\begingroup$ Perfect, this is the explanation I was looking for. So now after a few steps, I'll get ln(3/2) = ln(3) - ln(2). $\endgroup$ – Miguel Aragon Jul 11 at 21:36
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$$\begin{align} -2\ln \left( \frac{2}{\sqrt 6} \right) &= -2\big( \ln(2)-\ln(\sqrt{6}) \big) \\ &= -2\ln(2)+2\ln(6^{1/2}) \\ &= -2\ln(2)+\ln(2\cdot 3) \\ &=-2\ln(2)+ \big( \ln(2)+\ln(3)\big) \\ &=\ln(3)-\ln(2) \end{align}$$ And for your specific question, remember that $$\left( \frac{a}{b} \right)^{-n}=\left( \frac{b}{a} \right)^n$$

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Well you may start by distributing the index since $2$ and $\sqrt 6$ are positive. Thus $$\left(\frac{2}{\sqrt{6}}\right)^{-2}=\frac{2^{-2}}{{\sqrt 6}^{-2}}.$$

Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^{-n}=\frac {1}{a^n}.$$ Applying this to your expression, we have $$\frac{2^{-2}}{{\sqrt 6}^{-2}}=\frac{\frac {1}{2^2}}{\frac{1}{{\sqrt 6}^2}}=\frac{\frac {1}{4}}{\frac{1}{6}}=\frac{6}{4}=\frac 32.$$

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Using your first step,

$-2 \ln(\frac{2}{\sqrt{6}}) = \ln(\frac{2}{\sqrt{6}})^{-2} = \ln \frac{1}{(\frac{2}{\sqrt{6}})^{2}} = \ln \frac{6}{4} = \ln\frac{3}{2} = \ln 3 - \ln 2$

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$$ -2 \ln \left( \frac{2}{\sqrt{6}} \right) = \ln 3 - \ln 2$$

if and only if

$$ \ln \left( \frac{2}{\sqrt{6}} \right)^{-2} = \ln \left( \frac{3}{2}\right) $$

if and only if

$$ \ln \left[ \frac{1}{\left(\frac{2}{\sqrt{6}}\right)^{2}} \right] = \ln \left( \frac{3}{2}\right)$$

And so, we have

$$ \ln \left( \frac{6}{4} \right) = \ln \left( \frac{3}{2}\right)$$

which is true.

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  • $\begingroup$ you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse $\endgroup$ – peek-a-boo Jul 11 at 20:27
  • $\begingroup$ One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to include them initially. $\endgroup$ – mlchristians Jul 11 at 20:29
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    $\begingroup$ $4 = 10 \implies 7-3 =7+3 \implies-3= 3 \implies (-3)^2 = 3^2 \implies 9=9$ which is true. Therefore $4 = 10$. Or if penguins were elephants then they would both eat food. Which is true. So penguins are elephants. $\endgroup$ – fleablood Jul 11 at 20:37
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    $\begingroup$ "One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained." But if you do that you MUST specify the steps are actually if and only if steps. If you do not SPECIFICALLY state that it is reasonable to assume they are only forward implications and the proof is not valid. $\endgroup$ – fleablood Jul 11 at 20:38
  • $\begingroup$ Your first comment is not if and only if; and neither is you second. Forward implications is one-way; necessary and sufficient (i.e., if and only if are both ways). In the case of this problem, start from the given equation and work it to where I finished. Any problem with that. Now, if you want to do the work, start w/ the last statement and by a series of implications, show it implies the initial equation. I chose iff and only if instead. $\endgroup$ – mlchristians Jul 11 at 20:45
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For a non-zero real number, we have $$1=x^0=x^{2-2}=x^2\cdot x^{-2}\tag1$$ from How to understand why $x^0=1$ where $x$ is any real number?

Putting $x=2/\sqrt6$ into $(1)$, we get $$1=\left(\frac2{\sqrt6}\right)^2\cdot\left(\frac2{\sqrt6}\right)^{-2}\tag2$$ and since we know that for a positive integer $n$ and positive real numbers $a,b$, $$\left(\frac ab\right)^n=\underbrace{\frac ab\cdot \frac ab\cdot\ldots\cdot\frac ab}_{n\,\text{times}}=\underbrace{\frac{a\cdot a\cdot\ldots\cdot a}{b\cdot b\cdot\ldots\cdot b}}_{n\,\text{times}}=\frac{a^n}{b^n},\tag3$$ we have that $$\left(\frac2{\sqrt6}\right)^2=\frac{2^2}{\left(\sqrt6\right)^2}=\frac46=\frac23\tag4.$$ Finally, putting $(4)$ back into $(2)$ yields $$1=\frac23\cdot\left(\frac2{\sqrt6}\right)^{-2}\implies\left(\frac2{\sqrt6}\right)^{-2}=\frac1{2/3}=\frac32\tag5$$ as the answer.

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