Question

simplify fractional exponents (type 2)
score: 2/5 penalty: 1 off
question
simplify the following expression to simplest form using only positive exponents.
\\(\left(125x^{-9}y^{-21}\
ight)^{\frac{4}{3}}\\)
answer attempt 1 out of 2

Explanation:

Step1: Apply the power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). So we can apply this to \((125x^{-9}y^{-21})^{\frac{4}{3}}\) as follows:

$$ (125x^{-9}y^{-21})^{\frac{4}{3}} = 125^{\frac{4}{3}} \cdot (x^{-9})^{\frac{4}{3}} \cdot (y^{-21})^{\frac{4}{3}} $$

Step2: Simplify \(125^{\frac{4}{3}}\)

We know that \(125 = 5^3\), so we can rewrite \(125^{\frac{4}{3}}\) as \((5^3)^{\frac{4}{3}}\). Using the power of a power rule \((a^m)^n = a^{mn}\), we get:

$$ (5^3)^{\frac{4}{3}} = 5^{3\times\frac{4}{3}} = 5^4 = 625 $$

Step3: Simplify \((x^{-9})^{\frac{4}{3}}\)

Using the power of a power rule \((a^m)^n = a^{mn}\) again, we have:

$$ (x^{-9})^{\frac{4}{3}} = x^{-9\times\frac{4}{3}} = x^{-12} $$

Since we want only positive exponents, we know that \(x^{-n}=\frac{1}{x^n}\), so \(x^{-12} = \frac{1}{x^{12}}\)

Step4: Simplify \((y^{-21})^{\frac{4}{3}}\)

Using the power of a power rule \((a^m)^n = a^{mn}\) again, we have:

$$ (y^{-21})^{\frac{4}{3}} = y^{-21\times\frac{4}{3}} = y^{-28} $$

Since we want only positive exponents, we know that \(y^{-n}=\frac{1}{y^n}\), so \(y^{-28} = \frac{1}{y^{28}}\)

Step5: Combine the results

Now we combine the results from Step2, Step3, and Step4:

$$ 125^{\frac{4}{3}} \cdot (x^{-9})^{\frac{4}{3}} \cdot (y^{-21})^{\frac{4}{3}} = 625 \cdot \frac{1}{x^{12}} \cdot \frac{1}{y^{28}}=\frac{625}{x^{12}y^{28}} $$

Answer:

\(\frac{625}{x^{12}y^{28}}\)