Math worksheet on exponents and powers with problems for finding values, simplifying expressions, and expressing numbers in exponential form.

Problem Set: Exponents & Powers

Below, I will solve each problem step by step and explain the solution.

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Problem 1: Find the value of each of the following

#### (a) \( 13^2 \)
\[ 13^2 = 13 \times 13 = 169 \]

#### (b) \( 5^3 \)
\[ 5^3 = 5 \times 5 \times 5 = 125 \]

#### (c) \( 2^4 \)
\[ 2^4 = 2 \times 2 \times 2 \times 2 = 16 \]

#### (d) \( 11^2 \)
\[ 11^2 = 11 \times 11 = 121 \]

#### (e) \( (-3)^3 \)
\[ (-3)^3 = (-3) \times (-3) \times (-3) = -27 \]

#### (f) \( (-1)^6 \)
\[ (-1)^6 = (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) = 1 \]

Final Answers for Problem 1:
\[
\boxed{169, 125, 16, 121, -27, 1}
\]

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Problem 2: Simplify

#### (a) \( 3 \times 10^2 \)
\[ 3 \times 10^2 = 3 \times 100 = 300 \]

#### (b) \( 25 \times 5^3 \)
\[ 25 \times 5^3 = 25 \times (5 \times 5 \times 5) = 25 \times 125 = 3125 \]

#### (c) \( 0 \times 10^4 \)
\[ 0 \times 10^4 = 0 \]

#### (d) \( \left( \frac{3}{4} \right)^3 \)
\[ \left( \frac{3}{4} \right)^3 = \frac{3^3}{4^3} = \frac{27}{64} \]

#### (e) \( \left( \frac{-2}{3} \right)^4 \)
\[ \left( \frac{-2}{3} \right)^4 = \frac{(-2)^4}{3^4} = \frac{16}{81} \]

Final Answers for Problem 2:
\[
\boxed{300, 3125, 0, \frac{27}{64}, \frac{16}{81}}
\]

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Problem 3: Express each of the following in exponential form

#### (a) \( \left( \frac{-5}{7} \right) \times \left( \frac{-5}{7} \right) \times \left( \frac{-5}{7} \right) \times \left( \frac{-5}{7} \right) \)
This is a product of four identical terms:
\[ \left( \frac{-5}{7} \right)^4 \]

#### (b) \( -5 \times -5 \times -5 \)
This is a product of three identical terms:
\[ (-5)^3 \]

#### (c) \( x \times x \times x \times x \times a \times a \times b \times b \times b \)
Group the like terms:
\[ x^4 \times a^2 \times b^3 \]

#### (d) \( (-2) \times (-2) \times (-2) \times (-2) \times a \times a \times a \)
Group the like terms:
\[ (-2)^4 \times a^3 \]

Final Answers for Problem 3:
\[
\boxed{\left( \frac{-5}{7} \right)^4, (-5)^3, x^4 \cdot a^2 \cdot b^3, (-2)^4 \cdot a^3}
\]

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Problem 4: Express each of the following numbers as a product of powers of their prime factors

#### (a) \( 36 \)
Factorize \( 36 \):
\[ 36 = 2 \times 18 = 2 \times 2 \times 9 = 2^2 \times 3^2 \]

#### (b) \( 675 \)
Factorize \( 675 \):
\[ 675 = 3 \times 225 = 3 \times 3 \times 75 = 3 \times 3 \times 3 \times 25 = 3^3 \times 5^2 \]

#### (c) \( 392 \)
Factorize \( 392 \):
\[ 392 = 2 \times 196 = 2 \times 2 \times 98 = 2 \times 2 \times 2 \times 49 = 2^3 \times 7^2 \]

#### (d) \( 864 \)
Factorize \( 864 \):
\[ 864 = 2 \times 432 = 2 \times 2 \times 216 = 2 \times 2 \times 2 \times 108 = 2 \times 2 \times 2 \times 2 \times 54 = 2 \times 2 \times 2 \times 2 \times 2 \times 27 = 2^5 \times 3^3 \]

#### (e) \( 450 \)
Factorize \( 450 \):
\[ 450 = 2 \times 225 = 2 \times 3 \times 75 = 2 \times 3 \times 3 \times 25 = 2 \times 3^2 \times 5^2 \]

#### (f) \( 1800 \)
Factorize \( 1800 \):
\[ 1800 = 2 \times 900 = 2 \times 2 \times 450 = 2 \times 2 \times 2 \times 225 = 2 \times 2 \times 2 \times 3 \times 75 = 2 \times 2 \times 2 \times 3 \times 3 \times 25 = 2^3 \times 3^2 \times 5^2 \]

Final Answers for Problem 4:
\[
\boxed{2^2 \cdot 3^2, 3^3 \cdot 5^2, 2^3 \cdot 7^2, 2^5 \cdot 3^3, 2 \cdot 3^2 \cdot 5^2, 2^3 \cdot 3^2 \cdot 5^2}
\]

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Problem 5: Using laws of exponents, simplify

#### (i) \( 3^2 \times 3^5 \)
Using the law \( a^m \times a^n = a^{m+n} \):
\[ 3^2 \times 3^5 = 3^{2+5} = 3^7 \]

#### (ii) \( (7^2)^3 \div 7^2 \)
Using the laws \( (a^m)^n = a^{m \cdot n} \) and \( a^m \div a^n = a^{m-n} \):
\[ (7^2)^3 = 7^{2 \cdot 3} = 7^6 \]
\[ 7^6 \div 7^2 = 7^{6-2} = 7^4 \]

#### (iii) \( 2^{20} \div 2^5 \)
Using the law \( a^m \div a^n = a^{m-n} \):
\[ 2^{20} \div 2^5 = 2^{20-5} = 2^{15} \]

#### (iv) \( 2^4 \times 5^4 \)
Using the law \( a^m \times b^m = (a \cdot b)^m \):
\[ 2^4 \times 5^4 = (2 \cdot 5)^4 = 10^4 \]

#### (v) \( (2^0 + 3^0)(4^0 + 6^0) \)
Using the law \( a^0 = 1 \):
\[ 2^0 + 3^0 = 1 + 1 = 2 \]
\[ 4^0 + 6^0 = 1 + 1 = 2 \]
\[ (2^0 + 3^0)(4^0 + 6^0) = 2 \times 2 = 4 \]

#### (vi) \( \frac{7^2}{5^3} \)
This is already in its simplest form:
\[ \frac{7^2}{5^3} \]

Final Answers for Problem 5:
\[
\boxed{3^7, 7^4, 2^{15}, 10^4, 4, \frac{7^2}{5^3}}
\]

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Problem 6: Simplify and express each of the following in exponential form

#### (i) \( \frac{2^{15}}{2^3} \)
Using the law \( a^m \div a^n = a^{m-n} \):
\[ \frac{2^{15}}{2^3} = 2^{15-3} = 2^{12} \]

#### (ii) \( (3^5 \times 2^2)^3 \)
Using the laws \( (a \cdot b)^m = a^m \cdot b^m \) and \( (a^m)^n = a^{m \cdot n} \):
\[ (3^5 \times 2^2)^3 = (3^5)^3 \times (2^2)^3 = 3^{5 \cdot 3} \times 2^{2 \cdot 3} = 3^{15} \times 2^6 \]

#### (iii) \( [(2^3)^4 \times 2^8] + 2^{12} \)
Using the laws \( (a^m)^n = a^{m \cdot n} \) and \( a^m \times a^n = a^{m+n} \):
\[ (2^3)^4 = 2^{3 \cdot 4} = 2^{12} \]
\[ 2^{12} \times 2^8 = 2^{12+8} = 2^{20} \]
\[ 2^{20} + 2^{12} \]
Since the bases are the same but the exponents are different, this cannot be simplified further:
\[ 2^{20} + 2^{12} \]

#### (iv) \( \frac{5^4 \times x^{10} \times y^5}{3^4 \times x^7 \times y^4} \)
Using the laws \( \frac{a^m}{a^n} = a^{m-n} \) and \( a^m \times b^m = (a \cdot b)^m \):
\[ \frac{5^4}{3^4} = \left( \frac{5}{3} \right)^4 \]
\[ \frac{x^{10}}{x^7} = x^{10-7} = x^3 \]
\[ \frac{y^5}{y^4} = y^{5-4} = y^1 = y \]
Combining these:
\[ \frac{5^4 \times x^{10} \times y^5}{3^4 \times x^7 \times y^4} = \left( \frac{5}{3} \right)^4 \times x^3 \times y \]

#### (v) \( \left( \frac{2}{3} \right)^5 \times \left( \frac{3}{5} \right)^5 \)
Using the law \( \left( \frac{a}{b} \right)^m \times \left( \frac{c}{d} \right)^m = \left( \frac{a \cdot c}{b \cdot d} \right)^m \):
\[ \left( \frac{2}{3} \right)^5 \times \left( \frac{3}{5} \right)^5 = \left( \frac{2 \cdot 3}{3 \cdot 5} \right)^5 = \left( \frac{6}{15} \right)^5 = \left( \frac{2}{5} \right)^5 \]

#### (vi) \( \frac{9^8 \times (x^2)^5}{(27)^4 \times (x^3)^2} \)
Using the laws \( (a^m)^n = a^{m \cdot n} \) and \( \frac{a^m}{a^n} = a^{m-n} \):
\[ 9^8 = (3^2)^8 = 3^{16} \]
\[ (x^2)^5 = x^{2 \cdot 5} = x^{10} \]
\[ 27^4 = (3^3)^4 = 3^{12} \]
\[ (x^3)^2 = x^{3 \cdot 2} = x^6 \]
Substitute these into the expression:
\[ \frac{9^8 \times (x^2)^5}{(27)^4 \times (x^3)^2} = \frac{3^{16} \times x^{10}}{3^{12} \times x^6} \]
Simplify using the laws \( \frac{a^m}{a^n} = a^{m-n} \):
\[ \frac{3^{16}}{3^{12}} = 3^{16-12} = 3^4 \]
\[ \frac{x^{10}}{x^6} = x^{10-6} = x^4 \]
Combine these:
\[ \frac{3^{16} \times x^{10}}{3^{12} \times x^6} = 3^4 \times x^4 \]

Final Answers for Problem 6:
\[
\boxed{2^{12}, 3^{15} \cdot 2^6, 2^{20} + 2^{12}, \left( \frac{5}{3} \right)^4 \cdot x^3 \cdot y, \left( \frac{2}{5} \right)^5, 3^4 \cdot x^4}
\]

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Final Answer Summary:

\[
\boxed{
1. 169, 125, 16, 121, -27, 1 \\
2. 300, 3125, 0, \frac{27}{64}, \frac{16}{81} \\
3. \left( \frac{-5}{7} \right)^4, (-5)^3, x^4 \cdot a^2 \cdot b^3, (-2)^4 \cdot a^3 \\
4. 2^2 \cdot 3^2, 3^3 \cdot 5^2, 2^3 \cdot 7^2, 2^5 \cdot 3^3, 2 \cdot 3^2 \cdot 5^2, 2^3 \cdot 3^2 \cdot 5^2 \\
5. 3^7, 7^4, 2^{15}, 10^4, 4, \frac{7^2}{5^3} \\
6. 2^{12}, 3^{15} \cdot 2^6, 2^{20} + 2^{12}, \left( \frac{5}{3} \right)^4 \cdot x^3 \cdot y, \left( \frac{2}{5} \right)^5, 3^4 \cdot x^4
}
\]