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How do I figure this out:

$$\frac{27^{-2/3}}{16^{3/4}}$$

Please show workings. Thanks!

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closed as off-topic by Thomas, Brevan Ellefsen, iadvd, Claude Leibovici, naslundx Sep 1 '16 at 7:54

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  • $\begingroup$ Do you know what a fractional power means? It means taking a root. So $16^\frac{1}{4}$ would be $2$, because $2^4=16$. Can you use this logic to your benefit? $\endgroup$ – астон вілла олоф мэллбэрг Aug 31 '16 at 11:05
  • $\begingroup$ Ah so it would be -2(3)/3(2) after that step to both sides, meaning -6/6 equalling -1. Thanks so much!! $\endgroup$ – Kiwi Aug 31 '16 at 11:08
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    $\begingroup$ You've figured out the secret, but there's a mistake in your calculation. You will get $\frac{3^{-2}}{2^3}$, which will give you $\frac{1}{3^22^3}$, which gives you $\frac{1}{72}$. $\endgroup$ – астон вілла олоф мэллбэрг Aug 31 '16 at 11:10
  • $\begingroup$ Although you seem to refuse to use MathJax (e.g. from your other questions), I strongly suggest to format your questions and comments (note that your comments cannot be edited by others). May be then you could avoid such errors like writing $3^{-2}/2^3$ as $-2(3)/3(2) = -6/6 = -1.$ $\endgroup$ – gammatester Aug 31 '16 at 11:18
  • $\begingroup$ @gammatester Not exactly the best with MathJax, also I had thought that it was supposed to be multiplied not raised to the power with. $\endgroup$ – Kiwi Aug 31 '16 at 11:21
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Ok, so first we can see that both 27 and 16 can be written as powers of 3 and 2 respectively, so the expression in your question becomes $$\frac{(3^3)^{-\frac{2}{3}}}{(2^4)^{\frac{3}{4}}}$$ Next, we need to use a few index laws to simplify this. Remember that $(a^m)^n = a^{mn}$ and that $a^{-m} = \frac{1}{a^m}$. From this, we can simplify the above expression $$\frac{(3^3)^{-\frac{2}{3}}}{(2^4)^{\frac{3}{4}}} = \frac{3^{-2}}{2^{3}} = \frac{1}{3^22^3}=\frac{1}{72}$$

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Fractional exponents

$$27^{-2/3} = \frac{1}{27^{2/3}}$$

That is the first step, because you firstly have to remove the minus sign.

Now the fractional part:

$$27^{2/3} = \sqrt[3]{27^2}$$

Hence

$$\frac{27^{-2/3}}{16^{3/4}} = \frac{1}{27^{2/3} 16^{3/4}} = \frac{1}{\sqrt[3]{27^2} \sqrt[4]{16^3}}$$

Then, if you want, you can always arrange it better by writing

$$27 = 3^3 ~~~~~ 16 = 2^4$$

And have fun with exponents. Try!

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Hint: $$27^{-2/3} = \frac{1}{27^{2/3}} = \frac{1}{\sqrt[3]{27^2}}$$ and $$16^{3/4} = \sqrt[4]{16^3}$$

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    $\begingroup$ Somehow I feel $\left(\dfrac{1}{\sqrt[3]{27}}\right)^2$ and $\left(\sqrt[4]{16}\right)^3$ would be easier to do in my head $\endgroup$ – Henry Aug 31 '16 at 11:10
  • $\begingroup$ @Henry It's true too, your expression is a bit more developed. $\endgroup$ – Rubén Ballester Aug 31 '16 at 11:15
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Since $27^{-2/3}=\frac{1}{\sqrt[3]{27^2}}=\frac{1}{\sqrt[3]{3^{3^2}}}=\frac{1}{3^2}=\frac{1}{9}$ and $16^{3/4}=\sqrt[4]{16^3}=\sqrt[4]{2^{4^3}}=2^3$ you have

$$\frac{1}{9}\times\frac{1}{8}=\frac{1}{72}$$

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it is $$\frac{1}{16^{3/4}\cdot27^{2/3}}=\frac{1}{\sqrt[4]{16^3}\cdot\sqrt[3]{27^2}}$$ $$=\frac{1}{\sqrt[4]{2^{12}}\sqrt[3]{3^6}}=\frac{1}{2^33^2}=\frac{1}{8\cdot9}=\frac{1}{72}$$

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  • $\begingroup$ The $8$ should be in the denominator $\endgroup$ – Henry Aug 31 '16 at 11:14
  • $\begingroup$ what do you meant? $\endgroup$ – Dr. Sonnhard Graubner Aug 31 '16 at 11:15
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    $\begingroup$ You have misread the question. $\endgroup$ – PM 2Ring Aug 31 '16 at 11:16
  • $\begingroup$ yes or the question was just edited! $\endgroup$ – Dr. Sonnhard Graubner Aug 31 '16 at 11:17
  • $\begingroup$ Perhaps it was re-edited within the grace period. The source of the OP's original version does not have a negative exponent on 16. $\endgroup$ – PM 2Ring Aug 31 '16 at 11:22

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