Simplifying Rational Exponents - Kuta Software - Free Printable

Here's the step-by-step solution to all 12 problems on simplifying rational exponents. I’ll explain each one clearly.

---

Key Rules Used:

- \( (a^m)^n = a^{m \cdot n} \)
- \( a^{-m} = \frac{1}{a^m} \)
- \( a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m \)
- \( (ab)^n = a^n b^n \)
- When multiplying same bases: \( a^m \cdot a^n = a^{m+n} \)
- When dividing same bases: \( \frac{a^m}{a^n} = a^{m-n} \)

---

## Problems 1–8: Simplify

---

1) \( (a^4)^{\frac{1}{2}} \)

Apply power rule: multiply exponents
→ \( a^{4 \cdot \frac{1}{2}} = a^2 \)

✔ Answer: \( a^2 \)

---

2) \( (27p^6)^{\frac{2}{3}} \)

Break into parts: \( 27^{\frac{2}{3}} \cdot (p^6)^{\frac{2}{3}} \)

- \( 27 = 3^3 \), so \( (3^3)^{\frac{2}{3}} = 3^{3 \cdot \frac{2}{3}} = 3^2 = 9 \)
- \( (p^6)^{\frac{2}{3}} = p^{6 \cdot \frac{2}{3}} = p^4 \)

✔ Answer: \( 9p^4 \)

---

3) \( (25b^4)^{-1.5} \)

Note: \( -1.5 = -\frac{3}{2} \)

So, \( (25b^4)^{-\frac{3}{2}} = 25^{-\frac{3}{2}} \cdot (b^4)^{-\frac{3}{2}} \)

- \( 25 = 5^2 \), so \( (5^2)^{-\frac{3}{2}} = 5^{-3} = \frac{1}{125} \)
- \( (b^4)^{-\frac{3}{2}} = b^{-6} = \frac{1}{b^6} \)

Multiply: \( \frac{1}{125} \cdot \frac{1}{b^6} = \frac{1}{125b^6} \)

✔ Answer: \( \frac{1}{125b^6} \)

---

4) \( (64m^4)^{-\frac{1}{2}} \)

\( 64 = 8^2 \) or \( 4^3 \), but better as \( 64 = 2^6 \)

So: \( (2^6 m^4)^{-\frac{1}{2}} = 2^{6 \cdot (-\frac{1}{2})} \cdot m^{4 \cdot (-\frac{1}{2})} = 2^{-3} \cdot m^{-2} = \frac{1}{8} \cdot \frac{1}{m^2} = \frac{1}{8m^2} \)

✔ Answer: \( \frac{1}{8m^2} \)

---

5) \( (a^3)^{\frac{1}{6}} \)

Multiply exponents: \( a^{3 \cdot \frac{1}{6}} = a^{\frac{1}{2}} = \sqrt{a} \)

✔ Answer: \( \sqrt{a} \) or \( a^{1/2} \) — both acceptable unless specified otherwise.

---

6) \( (9r^2)^{0.5} \)

Note: \( 0.5 = \frac{1}{2} \)

So: \( (9r^2)^{\frac{1}{2}} = \sqrt{9r^2} = \sqrt{9} \cdot \sqrt{r^2} = 3|r| \)

But since we’re in algebra and often assume variables represent positive numbers unless stated, we can write:

✔ Answer: \( 3r \)

*(If domain includes negative r, answer is \( 3|r| \), but for this level, \( 3r \) is expected.)*

---

7) \( (81x^{12})^{1.25} \)

Note: \( 1.25 = \frac{5}{4} \)

So: \( (81x^{12})^{\frac{5}{4}} = 81^{\frac{5}{4}} \cdot x^{12 \cdot \frac{5}{4}} = 81^{\frac{5}{4}} \cdot x^{15} \)

Now simplify \( 81^{\frac{5}{4}} \):

\( 81 = 3^4 \), so \( (3^4)^{\frac{5}{4}} = 3^{5} = 243 \)

Thus: \( 243x^{15} \)

✔ Answer: \( 243x^{15} \)

---

8) \( (216r^9)^{\frac{1}{3}} \)

\( 216 = 6^3 \), so:

\( (6^3 r^9)^{\frac{1}{3}} = 6^{3 \cdot \frac{1}{3}} \cdot r^{9 \cdot \frac{1}{3}} = 6^1 \cdot r^3 = 6r^3 \)

✔ Answer: \( 6r^3 \)

---

## Problems 9–12: Simplify — Answer must have only positive exponents, no fractional exponents in denominator

---

9) \( 2m^{-7} \cdot 4m^{\frac{5}{2}} \cdot 4m^{-2} \)

First, combine coefficients: \( 2 \cdot 4 \cdot 4 = 32 \)

Now combine exponents of \( m \):
\( m^{-7 + \frac{5}{2} - 2} = m^{-9 + \frac{5}{2}} = m^{-\frac{18}{2} + \frac{5}{2}} = m^{-\frac{13}{2}} \)

So expression becomes: \( 32m^{-\frac{13}{2}} \)

We need positive exponents only, so rewrite:

→ \( \frac{32}{m^{\frac{13}{2}}} \)

But the problem says no fractional exponents in the denominator. So convert \( m^{\frac{13}{2}} = \sqrt{m^{13}} = m^6 \sqrt{m} \)

So: \( \frac{32}{m^6 \sqrt{m}} \)

Alternatively, write as \( \frac{32}{m^{6.5}} \), but that still has fractional exponent.

Better to write using radical form:

✔ Answer: \( \frac{32}{m^6 \sqrt{m}} \)

*(You could also write \( \frac{32}{\sqrt{m^{13}}} \), but factoring out perfect squares is cleaner.)*

---

10) \( 3b^{-\frac{1}{2}} \cdot b^{\frac{4}{3}} \)

Combine exponents: \( b^{-\frac{1}{2} + \frac{4}{3}} \)

Find common denominator: LCD of 2 and 3 is 6

→ \( -\frac{3}{6} + \frac{8}{6} = \frac{5}{6} \)

So: \( 3b^{\frac{5}{6}} \)

This has a fractional exponent in numerator, which is allowed. The restriction is “no fractional exponents in the denominator”.

✔ Answer: \( 3b^{\frac{5}{6}} \) — already satisfies conditions.

---

11) \( \left( p^{\frac{3}{7}} \right)^{-2} \)

Multiply exponents: \( p^{\frac{3}{7} \cdot (-2)} = p^{-\frac{6}{7}} \)

Convert to positive exponent: \( \frac{1}{p^{\frac{6}{7}}} \)

Now, remove fractional exponent from denominator:

→ \( \frac{1}{\sqrt[7]{p^6}} \)

✔ Answer: \( \frac{1}{\sqrt[7]{p^6}} \)

---

12) \( \left( \frac{1}{a^{\frac{1}{3}}} \right)^{\frac{3}{2}} \)

First, rewrite as: \( \left( a^{-\frac{1}{3}} \right)^{\frac{3}{2}} = a^{-\frac{1}{3} \cdot \frac{3}{2}} = a^{-\frac{1}{2}} \)

Convert to positive exponent: \( \frac{1}{a^{\frac{1}{2}}} = \frac{1}{\sqrt{a}} \)

✔ Answer: \( \frac{1}{\sqrt{a}} \)

---

## ✔ Final Answers Summary:

1) \( a^2 \)
2) \( 9p^4 \)
3) \( \frac{1}{125b^6} \)
4) \( \frac{1}{8m^2} \)
5) \( \sqrt{a} \) or \( a^{1/2} \)
6) \( 3r \)
7) \( 243x^{15} \)
8) \( 6r^3 \)
9) \( \frac{32}{m^6 \sqrt{m}} \)
10) \( 3b^{\frac{5}{6}} \)
11) \( \frac{1}{\sqrt[7]{p^6}} \)
12) \( \frac{1}{\sqrt{a}} \)

Let me know if you’d like these written with radicals instead of fractional exponents for all!