# Calculating $(100/9)^{-3/2}$

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

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

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

• Both ways give the same result. Sep 15, 2021 at 3:28
• General $(\frac{x}{y})^{-a}=(\frac{y}{x})^{a}$ Sep 15, 2021 at 3:29
• 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 $\displaystyle \bigg(\bigg(\frac{9}{100}\bigg)^{1/2}\bigg)^3$. Sep 15, 2021 at 3:29

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!)

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:

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

• You're not really addressing the OP's question here... Sep 15, 2021 at 3:33
• 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 ... " Sep 15, 2021 at 3:34