Math worksheet for converting exponential expressions to radical form.

Explanation:
We are asked to write each expression in radical form. That means turning expressions like $ a^{\frac{m}{n}} $ into roots.

The rule is:
$$
a^{\frac{m}{n}} = \sqrt[n]{a^m}
$$
or equivalently,
$$
a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m
$$
Both are correct, but usually we write it as the *n*th root of $ a^m $.

Let’s go one by one:

1) $ 7^{\frac{1}{2}} $
- Denominator is 2 → square root
- Numerator is 1 → power of 1 (so just 7 inside)
→ $ \sqrt{7} $

2) $ 4^{\frac{4}{3}} $
- Denominator 3 → cube root
- Numerator 4 → raise 4 to the 4th power inside
→ $ \sqrt[3]{4^4} $
But $ 4^4 = 256 $, so $ \sqrt[3]{256} $.
Alternatively, you could write $ \left(\sqrt[3]{4}\right)^4 $, but standard radical form is $ \sqrt[3]{4^4} $ or simplified if possible.
Note: $ 4 = 2^2 $, so $ 4^4 = (2^2)^4 = 2^8 = 256 $, and $ \sqrt[3]{256} $ doesn’t simplify nicely (since 256 = $ 2^8 $, and $ 2^8 = 2^{6+2} = (2^2)^3 \cdot 2^2 = 4^3 \cdot 4 $, so $ \sqrt[3]{256} = \sqrt[3]{4^3 \cdot 4} = 4\sqrt[3]{4} $).
Wait — let’s simplify properly:

$ 4^{\frac{4}{3}} = (2^2)^{\frac{4}{3}} = 2^{\frac{8}{3}} = \sqrt[3]{2^8} = \sqrt[3]{256} $, and since $ 256 = 64 \cdot 4 = 4^3 \cdot 4 $, then
$ \sqrt[3]{4^3 \cdot 4} = 4 \sqrt[3]{4} $.
So the simplified radical form is $ 4\sqrt[3]{4} $.
But unless simplification is required, $ \sqrt[3]{4^4} $ is acceptable. However, most teachers expect simplification when possible. Let's keep that in mind.

But for now, since the problem says “write in radical form”, not necessarily *simplified*, either is okay — but better to simplify if easy.

Let’s do all first, then decide.

3) $ 2^{\frac{5}{3}} $
→ cube root of $ 2^5 = 32 $, so $ \sqrt[3]{32} $.
Can we simplify? $ 32 = 8 \cdot 4 = 2^3 \cdot 2^2 $, so
$ \sqrt[3]{2^3 \cdot 2^2} = 2 \sqrt[3]{4} $.
So simplified: $ 2\sqrt[3]{4} $

4) $ 7^{\frac{4}{3}} $
→ $ \sqrt[3]{7^4} = \sqrt[3]{2401} $.
7⁴ = 7² × 7² = 49 × 49 = 2401.
Does 2401 have a perfect cube factor?
Check cubes: 1³=1, 2³=8, 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512, 9³=729, 10³=1000, 11³=1331, 12³=1728, 13³=2197, 14³=2744 > 2401.
So largest cube ≤ 2401 is 13³ = 2197, but 2401 ÷ 2197 is not integer.
2401 = 7⁴ = 7³ × 7, so
$ \sqrt[3]{7^3 \cdot 7} = 7 \sqrt[3]{7} $. ✔ That’s simplified.

5) $ 6^{\frac{3}{2}} $
→ square root of $ 6^3 = 216 $, so $ \sqrt{216} $.
Simplify: 216 = 36 × 6 = 6² × 6, so
$ \sqrt{6^2 \cdot 6} = 6\sqrt{6} $

6) $ 2^{\frac{1}{6}} $
→ sixth root of 2: $ \sqrt[6]{2} $. No simplification needed.

Now let’s list final simplified radical forms:

1) $ \sqrt{7} $
2) $ 4^{\frac{4}{3}} = \sqrt[3]{4^4} = \sqrt[3]{256} = 4\sqrt[3]{4} $
3) $ 2^{\frac{5}{3}} = \sqrt[3]{32} = 2\sqrt[3]{4} $
4) $ 7^{\frac{4}{3}} = \sqrt[3]{7^4} = 7\sqrt[3]{7} $
5) $ 6^{\frac{3}{2}} = \sqrt{216} = 6\sqrt{6} $
6) $ 2^{\frac{1}{6}} = \sqrt[6]{2} $

Double-check each:

1) $ (\sqrt{7})^2 = 7 $, and $ 7^{1/2} $ squared is 7 → correct.

2) $ (4\sqrt[3]{4})^3 = 4^3 \cdot 4 = 64 \cdot 4 = 256 = 4^4 $, and $ (4^{4/3})^3 = 4^4 = 256 $ → correct.

3) $ (2\sqrt[3]{4})^3 = 8 \cdot 4 = 32 = 2^5 $, and $ (2^{5/3})^3 = 2^5 = 32 $ → correct.

4) $ (7\sqrt[3]{7})^3 = 343 \cdot 7 = 2401 = 7^4 $ → correct.

5) $ (6\sqrt{6})^2 = 36 \cdot 6 = 216 = 6^3 $, and $ (6^{3/2})^2 = 6^3 = 216 $ → correct.

6) $ (\sqrt[6]{2})^6 = 2 $, and $ (2^{1/6})^6 = 2 $ → correct.

All good.

Final Answer:
1) $ \sqrt{7} $
2) $ 4\sqrt[3]{4} $
3) $ 2\sqrt[3]{4} $
4) $ 7\sqrt[3]{7} $
5) $ 6\sqrt{6} $
6) $ \sqrt[6]{2} $