Comprehensive practice sheet for 7th graders covering the laws of exponents and powers.

Problem Set: Exponents and Powers

Let's solve each problem step by step.

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#### (1) Find the value of the following:

- A) \( 8^4 \)
\[
8^4 = (2^3)^4 = 2^{3 \times 4} = 2^{12}
\]
Calculating \( 2^{12} \):
\[
2^{12} = 4096
\]

- B) \( (-8)^3 \)
\[
(-8)^3 = (-2^3)^3 = (-2)^{3 \times 3} = (-2)^9
\]
Since \((-2)^9\) is negative:
\[
(-2)^9 = -512
\]

- C) \( 9^4 \)
\[
9^4 = (3^2)^4 = 3^{2 \times 4} = 3^8
\]
Calculating \( 3^8 \):
\[
3^8 = 6561
\]

- D) \( (-7)^3 \)
\[
(-7)^3 = -(7^3)
\]
Calculating \( 7^3 \):
\[
7^3 = 343 \quad \Rightarrow \quad (-7)^3 = -343
\]

Final Answers:
\[
\boxed{4096, -512, 6561, -343}
\]

---

#### (2) Simplify the following:
\[
\left( \frac{-2}{2} \right)^2 \times \left( \frac{1}{2} \right)^3 \times \left( \frac{2}{3} \right)^3
\]

1. Simplify each term:
\[
\frac{-2}{2} = -1 \quad \Rightarrow \quad \left( \frac{-2}{2} \right)^2 = (-1)^2 = 1
\]
\[
\left( \frac{1}{2} \right)^3 = \frac{1^3}{2^3} = \frac{1}{8}
\]
\[
\left( \frac{2}{3} \right)^3 = \frac{2^3}{3^3} = \frac{8}{27}
\]

2. Multiply the results:
\[
1 \times \frac{1}{8} \times \frac{8}{27} = \frac{1 \times 1 \times 8}{8 \times 27} = \frac{8}{216} = \frac{1}{27}
\]

Final Answer:
\[
\boxed{\frac{1}{27}}
\]

---

#### (3) Simplify the following and write the answer in exponential form:

- A) \( 5^2 \times 5^4 \times 5^4 \times 5^7 \)
Using the property \( a^m \times a^n = a^{m+n} \):
\[
5^2 \times 5^4 \times 5^4 \times 5^7 = 5^{2+4+4+7} = 5^{17}
\]

- B) \( 2^5 \times 2^5 \)
\[
2^5 \times 2^5 = 2^{5+5} = 2^{10}
\]

- C) \( 3^6 \times 3^7 \times 3^5 \)
\[
3^6 \times 3^7 \times 3^5 = 3^{6+7+5} = 3^{18}
\]

- D) \( 3^9 \times 3^4 \times 3^3 + 3^8 \)
First, simplify the product:
\[
3^9 \times 3^4 \times 3^3 = 3^{9+4+3} = 3^{16}
\]
Then add \( 3^8 \):
\[
3^{16} + 3^8
\]
Since the bases are the same but the exponents are different, this cannot be simplified further in exponential form.

Final Answers:
\[
\boxed{5^{17}, 2^{10}, 3^{18}, 3^{16} + 3^8}
\]

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#### (4) If \( x = 1 \) and \( y = 5 \), find the value of \( \left( \frac{x}{y} \right)^5 \):

Substitute \( x = 1 \) and \( y = 5 \):
\[
\left( \frac{x}{y} \right)^5 = \left( \frac{1}{5} \right)^5 = \frac{1^5}{5^5} = \frac{1}{3125}
\]

Final Answer:
\[
\boxed{\frac{1}{3125}}
\]

---

#### (5) Find the number for the following expanded forms:

- A) \( 0 \times 10^2 + 0 \times 10^3 + 0 \times 10^4 + 5 \times 10^5 + 9 \times 10^6 \)
\[
0 \times 10^2 + 0 \times 10^3 + 0 \times 10^4 + 5 \times 10^5 + 9 \times 10^6 = 0 + 0 + 0 + 500000 + 9000000 = 9500000
\]

- B) \( 0 \times 10^1 + 9 \times 10^4 + 2 \times 10^2 + 1 \times 10^5 + 0 \times 10^7 \)
\[
0 \times 10^1 + 9 \times 10^4 + 2 \times 10^2 + 1 \times 10^5 + 0 \times 10^7 = 0 + 90000 + 200 + 100000 + 0 = 190200
\]

Final Answers:
\[
\boxed{9500000, 190200}
\]

---

#### (6) Simplify the following:
\[
\left( \frac{-5}{3} \right)^3 \times \left( \frac{1}{3} \right)^3
\]

1. Simplify each term:
\[
\left( \frac{-5}{3} \right)^3 = \frac{(-5)^3}{3^3} = \frac{-125}{27}
\]
\[
\left( \frac{1}{3} \right)^3 = \frac{1^3}{3^3} = \frac{1}{27}
\]

2. Multiply the results:
\[
\frac{-125}{27} \times \frac{1}{27} = \frac{-125 \times 1}{27 \times 27} = \frac{-125}{729}
\]

Final Answer:
\[
\boxed{\frac{-125}{729}}
\]

---

#### (7) If \( 2^p + 2^{p+1} = 24 \), find the value of \( p \):

1. Factor out \( 2^p \):
\[
2^p + 2^{p+1} = 2^p + 2^p \cdot 2 = 2^p (1 + 2) = 2^p \cdot 3
\]
So the equation becomes:
\[
2^p \cdot 3 = 24
\]

2. Solve for \( 2^p \):
\[
2^p = \frac{24}{3} = 8
\]

3. Express 8 as a power of 2:
\[
8 = 2^3 \quad \Rightarrow \quad 2^p = 2^3 \quad \Rightarrow \quad p = 3
\]

Final Answer:
\[
\boxed{3}
\]

---

#### (8) Choose the correct answer(s) from the given choice:
\[
\left( \frac{4}{2} \right)^4 \div \left( \frac{1}{2} \right)^3 = ?
\]

1. Simplify each term:
\[
\frac{4}{2} = 2 \quad \Rightarrow \quad \left( \frac{4}{2} \right)^4 = 2^4 = 16
\]
\[
\left( \frac{1}{2} \right)^3 = \frac{1^3}{2^3} = \frac{1}{8}
\]

2. Perform the division:
\[
\left( \frac{4}{2} \right)^4 \div \left( \frac{1}{2} \right)^3 = 16 \div \frac{1}{8} = 16 \times 8 = 128
\]

Final Answer:
\[
\boxed{128}
\]

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#### (9) Neha plants a jasmine on her 6th birthday. If the plant has one jasmine to start with, and the number of jasmines doubles every week, then how many jasmines will there be after \( x \) weeks?

The number of jasmines doubles every week. Starting with 1 jasmine:
\[
\text{After 1 week: } 1 \times 2 = 2
\]
\[
\text{After 2 weeks: } 1 \times 2^2 = 4
\]
\[
\text{After 3 weeks: } 1 \times 2^3 = 8
\]
\[
\text{After } x \text{ weeks: } 1 \times 2^x = 2^x
\]

Final Answer:
\[
\boxed{2^x}
\]

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Final Boxed Answers:

1. \(\boxed{4096, -512, 6561, -343}\)
2. \(\boxed{\frac{1}{27}}\)
3. \(\boxed{5^{17}, 2^{10}, 3^{18}, 3^{16} + 3^8}\)
4. \(\boxed{\frac{1}{3125}}\)
5. \(\boxed{9500000, 190200}\)
6. \(\boxed{\frac{-125}{729}}\)
7. \(\boxed{3}\)
8. \(\boxed{128}\)
9. \(\boxed{2^x}\)