Question

1. copy and complete the table.

	exponential form	radical form	evaluation of expression
a)	$36^{\frac{1}{2}}$	$\sqrt{36}$	6
b)	$16^{\frac{5}{2}}$	$\sqrt{16^5}$	32
c)	$1024^{\frac{1}{5}}$	$\sqrt5{1024}$	4
d)	$16807^{0.2}$
e)		$\sqrt3{-216^4}$	1296

Explanation:

Step 1: Recall Exponent-Radical Relationships

For any real number \(a\) and positive integers \(m, n\) (with \(n>1\)), \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\) or \((\sqrt[n]{a})^{m}\). Also, \(a^{-p}=\frac{1}{a^{p}}\) for \(a
eq0\) and \(p>0\), and \(\sqrt[n]{-a}=-\sqrt[n]{a}\) if \(n\) is odd.

Step 2: Solve Part (a)

• Exponential Form: \(36^{\frac{1}{2}}\)
• Radical Form: Using \(a^{\frac{1}{n}}=\sqrt[n]{a}\), here \(n = 2\), \(a=36\), so \(\sqrt{36}\)
• Evaluation: \(\sqrt{36}=6\) (since \(6\times6 = 36\))

Step 3: Solve Part (b)

• Exponential Form: \(16^{\frac{5}{2}}\)
• Radical Form: Using \(a^{\frac{m}{n}}=(\sqrt[n]{a})^{m}\), \(n = 2\), \(m = 5\), \(a = 16\), so \((\sqrt{16})^{5}\) or \(\sqrt{16^{5}}\). First, \(\sqrt{16}=4\), then \(4^{5}=1024\) (alternatively, \(16^{5}=(2^{4})^{5}=2^{20}\), \(\sqrt{2^{20}}=2^{10}=1024\))
• Evaluation: \(16^{\frac{5}{2}}=(\sqrt{16})^{5}=4^{5}=1024\)? Wait, wait, the user wrote 32? Wait, maybe a typo. Wait, \(16^{\frac{5}{2}}\): \(\sqrt{16}=4\), \(4^{5}=1024\). If it was \(16^{\frac{5}{4}}\), then \(\sqrt[4]{16^{5}}=\sqrt[4]{(2^{4})^{5}}=2^{5}=32\). Maybe the exponent is \(\frac{5}{4}\) instead of \(\frac{5}{2}\). Let's assume the exponent is \(\frac{5}{4}\) (since the evaluation is 32). Then:
• Exponential Form: \(16^{\frac{5}{4}}\)
• Radical Form: \(\sqrt[4]{16^{5}}\) or \((\sqrt[4]{16})^{5}\)
• Evaluation: \(\sqrt[4]{16}=2\), \(2^{5}=32\) (since \(16 = 2^{4}\), so \(\sqrt[4]{16}=2\), \(2^{5}=32\))

Step 4: Solve Part (c)

• Exponential Form: Let the radical be \(\sqrt[5]{1024}\). Let \(a^{\frac{1}{5}}=\sqrt[5]{1024}\), so \(a = 1024^{\frac{1}{1}}\) wait, no. Wait, \(\sqrt[5]{1024}\): \(1024=2^{10}\), so \(\sqrt[5]{2^{10}}=2^{2}=4\). So the exponential form is \(1024^{\frac{1}{5}}\) (since \(\sqrt[5]{1024}=1024^{\frac{1}{5}}\))
• Radical Form: \(\sqrt[5]{1024}\)
• Evaluation: \(1024^{\frac{1}{5}}=\sqrt[5]{1024}=4\) (since \(4^{5}=1024\))

Step 5: Solve Part (d)

• Exponential Form: \(16807^{-0.2}\). First, \(0.2=\frac{1}{5}\), so \(16807^{-0.2}=16807^{-\frac{1}{5}}\). Using \(a^{-p}=\frac{1}{a^{p}}\), this is \(\frac{1}{16807^{\frac{1}{5}}}\). Now, \(16807 = 7^{5}\) (since \(7\times7 = 49\), \(49\times7 = 343\), \(343\times7 = 2401\), \(2401\times7 = 16807\)), so \(16807^{\frac{1}{5}}=7\), thus \(16807^{-\frac{1}{5}}=\frac{1}{7}\)
• Radical Form: \(\frac{1}{\sqrt[5]{16807}}\) (since \(a^{-\frac{1}{n}}=\frac{1}{\sqrt[n]{a}}\))
• Evaluation: \(\frac{1}{7}\)

Step 6: Solve Part (e)

• Radical Form: \(\sqrt[3]{-216^{4}}\) (wait, the radical is \(\sqrt[3]{-216^{4}}\)? Wait, the user wrote \(\sqrt[3]{-216^{4}}\)? Wait, no, maybe \(\sqrt[3]{-216}\) raised to 4? Wait, the radical is \(\sqrt[3]{-216^{4}}\) or \((\sqrt[3]{-216})^{4}\). Let's check: \(\sqrt[3]{-216}=-6\) (since \((-6)^{3}=-216\)), then \((-6)^{4}=1296\) (since even exponent makes it positive). So:
• Exponential Form: \((-216)^{\frac{4}{3}}\) (since \(\sqrt[3]{-216^{4}}=(-216)^{\frac{4}{3}}\) or \((\sqrt[3]{-216})^{4}=(-6)^{4}\))
• Radical Form: \((\sqrt[3]{-216})^{4}\) or \(\sqrt[3]{(-216)^{4}}\)
• Evaluation: \(\sqrt[3]{-216}=-6\), \((-6)^{4}=1296\) (the user wrote 1246, which is a typo; correct is 1296)

Answer:

(Completed Table):

Exponential Form	Radical Form	Evaluation of Expression
b) \(16^{\frac{5}{4}}\) (assuming exponent \(\frac{5}{4}\) for evaluation \(32\))	\(\sqrt[4]{16^{5}}\) or \((\sqrt[4]{16})^{5}\)	\(32\)
c) \(1024^{\frac{1}{5}}\)	\(\sqrt[5]{1024}\)	\(4\)
d) \(16807^{-\frac{1}{5}}\) (or \(16807^{-0.2}\))	\(\frac{1}{\sqrt[5]{16807}}\)	\(\frac{1}{7}\)
e) \((-216)^{\frac{4}{3}}\)	\((\sqrt[3]{-216})^{4}\)	\(1296\) (corrected from 1246)

(Note: For part (b), if the intended exponent was \(\frac{5}{2}\), the evaluation would be \(1024\) instead of \(32\); the discrepancy suggests a possible typo in the exponent, likely \(\frac{5}{4}\) to match the evaluation \(32\).)