Concept-HW-G8- Laws of Exponents worksheet - Free Printable

To solve the problem, we need to simplify each expression on the left side using the laws of exponents and then match them with the corresponding options on the right side. Let's go through each expression step by step.

Expression 1: \( (-2)^7 \times (-2)^3 \)

Using the law of exponents \( a^m \times a^n = a^{m+n} \):

\[
(-2)^7 \times (-2)^3 = (-2)^{7+3} = (-2)^{10}
\]

Match: C. \( (-2)^{10} \)

---

Expression 2: \( a^3 \div a^5 \)

Using the law of exponents \( \frac{a^m}{a^n} = a^{m-n} \):

\[
a^3 \div a^5 = a^{3-5} = a^{-2}
\]

Match: I. \( a^{-2} \)

---

Expression 3: \( [(-3)^4]^2 \)

Using the law of exponents \( (a^m)^n = a^{m \cdot n} \):

\[
[(-3)^4]^2 = (-3)^{4 \cdot 2} = (-3)^8
\]

Match: H. \( (-3)^8 \)

---

Expression 4: \( 5^3 \times 5^6 \)

Using the law of exponents \( a^m \times a^n = a^{m+n} \):

\[
5^3 \times 5^6 = 5^{3+6} = 5^9
\]

Match: F. \( 5^9 \)

---

Expression 5: \( 13^0 \)

Using the law of exponents \( a^0 = 1 \) for any \( a \neq 0 \):

\[
13^0 = 1
\]

Match: A. \( 1 \)

---

Expression 6: \( 2^7 \div 2^4 \)

Using the law of exponents \( \frac{a^m}{a^n} = a^{m-n} \):

\[
2^7 \div 2^4 = 2^{7-4} = 2^3
\]

Match: D. \( 2^3 \)

---

Expression 7: \( \left(\frac{4}{9}\right)^3 \times \left(\frac{4}{9}\right)^2 \)

Using the law of exponents \( a^m \times a^n = a^{m+n} \):

\[
\left(\frac{4}{9}\right)^3 \times \left(\frac{4}{9}\right)^2 = \left(\frac{4}{9}\right)^{3+2} = \left(\frac{4}{9}\right)^5
\]

However, the options do not include \( \left(\frac{4}{9}\right)^5 \). Upon rechecking, it seems there might be a typo in the problem statement or options. Assuming the correct option is \( \left(\frac{4}{9}\right)^3 \times \left(\frac{4}{9}\right)^2 = \left(\frac{4}{9}\right)^5 \), but since it's not listed, we'll proceed with the given options.

Match: E. \( \left(\frac{4}{9}\right) \) (closest match, though not exact)

---

Expression 8: \( 7^8 \times 7^3 \div 7^4 \)

First, simplify \( 7^8 \times 7^3 \) using \( a^m \times a^n = a^{m+n} \):

\[
7^8 \times 7^3 = 7^{8+3} = 7^{11}
\]

Then, divide by \( 7^4 \) using \( \frac{a^m}{a^n} = a^{m-n} \):

\[
7^{11} \div 7^4 = 7^{11-4} = 7^7
\]

Match: B. \( 7^7 \)

---

Expression 9: \( (4^{-2})^3 \)

Using the law of exponents \( (a^m)^n = a^{m \cdot n} \):

\[
(4^{-2})^3 = 4^{-2 \cdot 3} = 4^{-6}
\]

Match: G. \( 4^{-6} \)

---

Final Matches:

1. \( (-2)^7 \times (-2)^3 \) → C. \( (-2)^{10} \)
2. \( a^3 \div a^5 \) → I. \( a^{-2} \)
3. \( [(-3)^4]^2 \) → H. \( (-3)^8 \)
4. \( 5^3 \times 5^6 \) → F. \( 5^9 \)
5. \( 13^0 \) → A. \( 1 \)
6. \( 2^7 \div 2^4 \) → D. \( 2^3 \)
7. \( \left(\frac{4}{9}\right)^3 \times \left(\frac{4}{9}\right)^2 \) → E. \( \left(\frac{4}{9}\right) \) (closest match)
8. \( 7^8 \times 7^3 \div 7^4 \) → B. \( 7^7 \)
9. \( (4^{-2})^3 \) → G. \( 4^{-6} \)

Boxed Answer:

\[
\boxed{
\begin{aligned}
1 &\rightarrow \text{C} \\
2 &\rightarrow \text{I} \\
3 &\rightarrow \text{H} \\
4 &\rightarrow \text{F} \\
5 &\rightarrow \text{A} \\
6 &\rightarrow \text{D} \\
7 &\rightarrow \text{E} \\
8 &\rightarrow \text{B} \\
9 &\rightarrow \text{G}
\end{aligned}
}
\]