Question

in exercises 19 and 20, rewrite the expression in rational exponent form. 19. $(sqrt3{8})^4$ 20. $(sqrt5{-21})^6$

Explanation:

Exercise 19:

Step1: Recall the radical to exponent rule

The rule is $\sqrt[n]{a^m}=a^{\frac{m}{n}}$, and $(\sqrt[n]{a})^m = a^{\frac{m}{n}}$. Here, $n = 3$, $a = 8$, $m = 4$.

Step2: Apply the rule

So $(\sqrt[3]{8})^4=8^{\frac{4}{3}}$. We can also simplify $8^{\frac{4}{3}}$ further. Since $8 = 2^3$, then $(2^3)^{\frac{4}{3}}$.

Step3: Simplify the exponent

Using the power - of - a - power rule $(a^m)^n=a^{m\times n}$, we have $2^{3\times\frac{4}{3}}=2^4 = 16$. But the question just asks for rational exponent form, so the first form is also correct.

Step1: Recall the radical to exponent rule

The rule is $(\sqrt[n]{a})^m=a^{\frac{m}{n}}$. Here, $n = 5$, $a=- 21$, $m = 6$.

Step2: Apply the rule

So $(\sqrt[5]{-21})^6=(-21)^{\frac{6}{5}}$. We can also rewrite $(-21)^{\frac{6}{5}}$ as $((-21)^{\frac{1}{5}})^6$ or note that $(-21)^{\frac{6}{5}}=\sqrt[5]{(-21)^6}=\sqrt[5]{21^6}$ (since the exponent 6 is even, $(-21)^6 = 21^6$), but the rational exponent form is $(-21)^{\frac{6}{5}}$.

Answer:

$8^{\frac{4}{3}}$ (or $16$ if simplified, but rational exponent form is $8^{\frac{4}{3}}$)

Exercise 20: