Question

27^-4/3

How do I solve this?

Answer 1

\[\huge\rm x^{-m} = \frac{ 1 }{ x^m }\]
if there is a negative exponent then you should flip the fraction

Answer 2

@Nnesha okay- what do I do after that?

Answer 3

\[27^{-\frac{ 4 }{ 3 }} = \frac{ 1 }{ 27^\frac{ 4 }{ 3 } } = \frac{ 1 }{ \left( 27^\frac{ 1 }{ 3 } \right)^4 }\]

Answer 4

And\[x^\frac{ 1 }{ 3 } = \sqrt[3]{x}\]

Answer 5

I'm a bit confused with that second equation. How/when would I use that?

Answer 6

Well, any time you have a fractional exponent, the denominator of the fraction is the root that is required. In other words,\[x^\frac{ n }{ m } = \sqrt[m]{x^n}\]And you have a fractional exponent in your question.

Answer 7

Put simply, you can simplify \(27^\frac{1}{3}\)

Answer 8

So does -4/3 get plugged into that or does 1/3

Answer 9

I already converted my equation

Answer 10

I'm sorry. I don't understand. Are you able to simplify\[\frac{ 1 }{ \left( 27^\frac{ 1 }{ 3 } \right)^4 }\]

Answer 11

I don't think so, but I'm asking if I use the exponent -4/3 or the exponent 1/3 to plug into that rational expression

Answer 12

This problem involves two exponent rules.
1. a negative exponent
2. a fractional exponent

Here are the rules you need:

Negative exponent:
\(\Large a ^{-n} = \dfrac{1}{a^n} \)

Fractional exponent:
\(\Large a^{\frac{m}{n}} = \sqrt[n] {a^m} = \left( \sqrt[n] a \right)^m \)