How do I figure this out:


Please show workings. Thanks!


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Thomas, Brevan Ellefsen, iadvd, Claude Leibovici, naslundx
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $$\frac{(3^3)^{-\frac{2}{3}}}{(2^4)^{\frac{3}{4}}} = \frac{3^{-2}}{2^{3}} = \frac{1}{3^22^3}=\frac{1}{72}$$


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:

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


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

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

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

And have fun with exponents. Try!


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

  • 4
    $\begingroup$ 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

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



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

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.