How do I take a fraction to a negative power?

Question

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

Answer 1

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

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

Answer 2

$$\begin{align} -2\ln \left( \frac{2}{\sqrt 6} \right) &= -2\big( \ln(2)-\ln(\sqrt{6}) \big) \\ &= -2\ln(2)+2\ln(6^{1/2}) \\ &= -2\ln(2)+\ln(2\cdot 3) \\ &=-2\ln(2)+ \big( \ln(2)+\ln(3)\big) \\ &=\ln(3)-\ln(2) \end{align}$$ And for your specific question, remember that $$\left( \frac{a}{b} \right)^{-n}=\left( \frac{b}{a} \right)^n$$

Answer 3

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 $$\frac{2^{-2}}{{\sqrt 6}^{-2}}=\frac{\frac {1}{2^2}}{\frac{1}{{\sqrt 6}^2}}=\frac{\frac {1}{4}}{\frac{1}{6}}=\frac{6}{4}=\frac 32.$$

Answer 4

Using your first step,

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

Answer 5

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

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.

Answer 6

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.