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$$\left( \frac{100}{9} \right)^{-3/2} = \frac{27}{1000}$$

I am familiar with and have a grasp on calculating exponents but this one right here.....man listen lol. Being how the exponent is negative would I flip the numerator and the denominator to get

$$\left(\frac{9}{100}\right)^{3/2}$$

or would I place the entire equation under $1$ as a numerator to get

$$\dfrac{1} {\left( \frac{ 100 }{9} \right)^{3/2}}$$

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    $\begingroup$ Both ways give the same result. $\endgroup$
    – John Douma
    Sep 15, 2021 at 3:28
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    $\begingroup$ General $(\frac{x}{y})^{-a}=(\frac{y}{x})^{a}$ $\endgroup$ Sep 15, 2021 at 3:29
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    $\begingroup$ Writing it as $\displaystyle \bigg(\frac{9}{100}\bigg)^{3/2}$ is the right idea (the other way is fine too but has the added complication of dividing by a fraction, so you'll have to multiply by the reciprocal). Note that this expression is the same as $\displaystyle \bigg(\bigg(\frac{9}{100}\bigg)^{1/2}\bigg)^3$. $\endgroup$ Sep 15, 2021 at 3:29

2 Answers 2

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Note that

$$x^{-y} = \frac{1}{x^y}$$

whenever this quantity is defined. In your case we have $y=3/2$ and $x=100/9$. Therefore,

$$\left( \frac{100}{9} \right)^{-3/2} = \frac{1}{(100/9)^{3/2}}$$

You can then split this fraction up, because

$$\frac{1}{a/b} = \frac{b}{a}$$

In this case, $a=100^{3/2}$ and $b=9^{3/2}$. Thus,

$$\left( \frac{100}{9} \right)^{-3/2} = \frac{1}{(100/9)^{3/2}} = \frac{9^{3/2}}{100^{3/2}}$$

Of course, if you want to pull out that power of $3/2$ and use it on the entire fraction, that works too:

$$\left( \frac{100}{9} \right)^{-3/2} = \frac{1}{(100/9)^{3/2}} = \frac{9^{3/2}}{100^{3/2}} = \left( \frac{9}{100} \right)^{3/2}$$


An alternate way to frame this is that

$$(x^y)^z = x^{yz} = (x^z)^y$$

Thus,

$$\left( \frac{100}{9} \right)^{-3/2} = \left( \left( \frac{100}{9} \right)^{-1} \right)^{3/2} $$

Then, since $x^{-1} = 1/x$, we flip the inside

$$\left( \frac{100}{9} \right)^{-3/2} = \left( \left( \frac{100}{9} \right)^{-1} \right)^{3/2} = \left( \frac{9}{100} \right)^{3/2} $$


In short, either method works. (Just be sure you know why each works!)

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Take the reciprocal and negate the exponent.

$$\left(\dfrac{100}{9}\right)^{-3/2} = \left(\dfrac{9}{100}\right)^{3/2} $$

Rewrite as $a^{m/n} = (\sqrt[n]{a})^m$:

$$ \left(\dfrac{9}{100}\right)^{3/2} = \left(\sqrt{\dfrac{9}{100}}\right)^3$$

Take the root, and raise it to the power:

$$\left(\sqrt{\dfrac{9}{100}}\right)^3 = \left(\dfrac{3}{10}\right)^3 = \dfrac{3^3}{10^3}$$

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  • $\begingroup$ You're not really addressing the OP's question here... $\endgroup$ Sep 15, 2021 at 3:33
  • $\begingroup$ I did. I didn't say yes, so I'll say it now, but one of OP's two options was "would I flip the numerator and the denominator ... " $\endgroup$
    – David P
    Sep 15, 2021 at 3:34

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