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$\left( \frac23 \right) ^{-2}$ I noticed that the answer is $\frac94$, and I can't come to the conclusion of why it is?

JMoravitz
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Jorden Danczak
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  • The verb "solve" is inappropriate here. We are not solving anything. We are evaluating or simplifying or expressing differently. – JMoravitz Sep 15 '17 at 19:55
  • You need to keep in mind the following : $$\forall x,y\in\mathbb{R}, x^{-y}={1\over x^y}$$ Now you can substitute $x$ and $y$ with your values and you'll find the answer =) – Furrane Sep 15 '17 at 19:59

3 Answers3

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Guide:

Approach $1$:

Well, first evaluate $\left( \frac23\right)^{-1}= \frac{a}{b}$

After which you can compute $\left(\frac{a}{b}\right)^2=\frac{a^2}{b^2}$

Approach $2$:

Well, first evaluate $\left( \frac23\right)^{2}= \frac{c}{d}$

After which you can compute $\left(\frac{c}{d}\right)^{-1}=\frac{d}{c}$

Siong Thye Goh
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$$x^{-1}=\frac 1x$$

Thus $$\left(\frac 23 \right)^{-2}=\left(\left(\frac 23\right)^2\right)^{-1}=\left(\frac 49 \right)^{-1}=\frac 94$$

Jaideep Khare
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Because $\;\displaystyle\Bigl(\frac{a}{b}\Bigr)^{-2}=\biggl(\Bigl(\frac{a}{b}\Bigr)^{-1}\biggr)^2=\Bigl(\frac{b}{a}\Bigr)^2=\frac{b^2}{a^2}.$

Bernard
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