Math worksheet focusing on exponent rules and operations.

Problem Solution

Let's solve each part of the problem step by step.

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1. Complete the following.

#### (a) The value of \(3^4 = \ldots\)
\[ 3^4 = 3 \times 3 \times 3 \times 3 = 81 \]

#### (b) \(125^0 + 5^2 = \ldots\)
- Any number raised to the power of 0 is 1: \(125^0 = 1\).
- \(5^2 = 5 \times 5 = 25\).
- Therefore, \(125^0 + 5^2 = 1 + 25 = 26\).

#### (c) Multiplicative inverse of \(2^{-7}\)
- The multiplicative inverse of a number \(a\) is \( \frac{1}{a} \).
- The multiplicative inverse of \(2^{-7}\) is:
\[
\frac{1}{2^{-7}} = 2^7
\]
- Calculate \(2^7\):
\[
2^7 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 128
\]

#### (d) \(11^4 \times 11^{-2} = \ldots\)
- Use the property of exponents: \(a^m \times a^n = a^{m+n}\).
- Here, \(11^4 \times 11^{-2} = 11^{4 + (-2)} = 11^2\).
- Calculate \(11^2\):
\[
11^2 = 11 \times 11 = 121
\]

#### (e) \(\frac{1}{7^{-4}} = \ldots\)
- Use the property of exponents: \(\frac{1}{a^{-n}} = a^n\).
- Therefore, \(\frac{1}{7^{-4}} = 7^4\).
- Calculate \(7^4\):
\[
7^4 = 7 \times 7 \times 7 \times 7 = 2401
\]

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2. Write 1723.56 in expanded form using exponents.

- Break down the number into its place values:
\[
1723.56 = 1 \times 10^3 + 7 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 + 5 \times 10^{-1} + 6 \times 10^{-2}
\]

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3. Find the value of \(4^3 \times 2^5 \times 2^{-6}\) in exponents.

- Simplify the expression step by step:
\[
4^3 \times 2^5 \times 2^{-6}
\]
- Note that \(4 = 2^2\), so \(4^3 = (2^2)^3 = 2^{2 \times 3} = 2^6\).
- Substitute \(4^3\) with \(2^6\):
\[
4^3 \times 2^5 \times 2^{-6} = 2^6 \times 2^5 \times 2^{-6}
\]
- Use the property of exponents \(a^m \times a^n = a^{m+n}\):
\[
2^6 \times 2^5 \times 2^{-6} = 2^{6+5+(-6)} = 2^{5}
\]
- Calculate \(2^5\):
\[
2^5 = 32
\]

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4. Simplify and express the result in power notation with positive exponents.

#### (a) \(\left(\frac{1}{3}\right)^{-3} \times \left(\frac{1}{3}\right)^4 \times (3)^{-2}\)
- Use the property \(\left(\frac{1}{a}\right)^n = a^{-n}\):
\[
\left(\frac{1}{3}\right)^{-3} = 3^3, \quad \left(\frac{1}{3}\right)^4 = 3^{-4}, \quad (3)^{-2} = 3^{-2}
\]
- Substitute these into the expression:
\[
\left(\frac{1}{3}\right)^{-3} \times \left(\frac{1}{3}\right)^4 \times (3)^{-2} = 3^3 \times 3^{-4} \times 3^{-2}
\]
- Use the property \(a^m \times a^n = a^{m+n}\):
\[
3^3 \times 3^{-4} \times 3^{-2} = 3^{3 + (-4) + (-2)} = 3^{-3}
\]

#### (b) \(\left(\frac{3}{7}\right)^4 \times \left(\frac{7}{3}\right)^2 \times \left(\frac{1}{7}\right)^{-2}\)
- Use the property \(\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}\):
\[
\left(\frac{3}{7}\right)^4 = \frac{3^4}{7^4}, \quad \left(\frac{7}{3}\right)^2 = \frac{7^2}{3^2}, \quad \left(\frac{1}{7}\right)^{-2} = 7^2
\]
- Substitute these into the expression:
\[
\left(\frac{3}{7}\right)^4 \times \left(\frac{7}{3}\right)^2 \times \left(\frac{1}{7}\right)^{-2} = \frac{3^4}{7^4} \times \frac{7^2}{3^2} \times 7^2
\]
- Combine the fractions:
\[
\frac{3^4}{7^4} \times \frac{7^2}{3^2} \times 7^2 = \frac{3^4 \cdot 7^2 \cdot 7^2}{7^4 \cdot 3^2}
\]
- Simplify the powers:
\[
\frac{3^4 \cdot 7^4}{7^4 \cdot 3^2} = \frac{3^{4-2} \cdot 7^{4-4}}{1} = 3^2 \cdot 7^0 = 3^2 \cdot 1 = 3^2
\]
- Calculate \(3^2\):
\[
3^2 = 9
\]

#### (c) \((5^{-2} \times 5^{-9}) \times 5^4\)
- Use the property \(a^m \times a^n = a^{m+n}\):
\[
5^{-2} \times 5^{-9} = 5^{-2 + (-9)} = 5^{-11}
\]
- Substitute back into the expression:
\[
(5^{-2} \times 5^{-9}) \times 5^4 = 5^{-11} \times 5^4
\]
- Simplify further:
\[
5^{-11} \times 5^4 = 5^{-11 + 4} = 5^{-7}
\]

#### (d) \((13^2 \times 4^{-2}) \times 8^2\)
- Simplify each term:
\[
13^2 = 169, \quad 4^{-2} = \frac{1}{4^2} = \frac{1}{16}, \quad 8^2 = 64
\]
- Substitute these into the expression:
\[
(13^2 \times 4^{-2}) \times 8^2 = (169 \times \frac{1}{16}) \times 64
\]
- Simplify step by step:
\[
169 \times \frac{1}{16} = \frac{169}{16}
\]
\[
\left(\frac{169}{16}\right) \times 64 = \frac{169 \times 64}{16}
\]
- Simplify the fraction:
\[
\frac{169 \times 64}{16} = 169 \times 4 = 676
\]

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5. Simplify: \((3)^{-2} \times \left(\frac{1}{3}\right)^2 \times \left(\frac{1}{3}\right)^{-3}\)

- Use the property \(\left(\frac{1}{a}\right)^n = a^{-n}\):
\[
\left(\frac{1}{3}\right)^2 = 3^{-2}, \quad \left(\frac{1}{3}\right)^{-3} = 3^3
\]
- Substitute these into the expression:
\[
(3)^{-2} \times \left(\frac{1}{3}\right)^2 \times \left(\frac{1}{3}\right)^{-3} = 3^{-2} \times 3^{-2} \times 3^3
\]
- Use the property \(a^m \times a^n = a^{m+n}\):
\[
3^{-2} \times 3^{-2} \times 3^3 = 3^{-2 + (-2) + 3} = 3^{-1}
\]
- Express \(3^{-1}\) as a fraction:
\[
3^{-1} = \frac{1}{3}
\]

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6. Write 'True' or 'False' for the following statements.

#### (a) \(\left(\frac{3}{11}\right)^{-2}\) is a whole number.
- Use the property \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\):
\[
\left(\frac{3}{11}\right)^{-2} = \left(\frac{11}{3}\right)^2 = \frac{11^2}{3^2} = \frac{121}{9}
\]
- \(\frac{121}{9}\) is not a whole number.
- Answer: False

#### (b) \(\left(\frac{2}{9}\right)^{-2} \times \left(\frac{9}{2}\right)^2 = 1\)
- Use the property \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\):
\[
\left(\frac{2}{9}\right)^{-2} = \left(\frac{9}{2}\right)^2
\]
- Substitute this into the expression:
\[
\left(\frac{2}{9}\right)^{-2} \times \left(\frac{9}{2}\right)^2 = \left(\frac{9}{2}\right)^2 \times \left(\frac{9}{2}\right)^2
\]
- Simplify:
\[
\left(\frac{9}{2}\right)^2 \times \left(\frac{9}{2}\right)^2 = \left(\frac{9}{2}\right)^{2+2} = \left(\frac{9}{2}\right)^4
\]
- \(\left(\frac{9}{2}\right)^4 \neq 1\).
- Answer: False

#### (c) \(\left[(a)^{-m}\right]^n = a^{mn}\)
- Use the property \((a^m)^n = a^{m \cdot n}\):
\[
\left[(a)^{-m}\right]^n = a^{-m \cdot n} = a^{-mn}
\]
- The given statement is incorrect because it should be \(a^{-mn}\), not \(a^{mn}\).
- Answer: False

#### (d) \(327900000 = (3.279) \times 10^8\)
- Convert \(327900000\) to scientific notation:
\[
327900000 = 3.279 \times 10^8
\]
- The statement is correct.
- Answer: True

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7. Simplify: \(\left[\left(\frac{2}{7}\right)^{-2}\right]^4 \times \left[\left(\frac{7}{2}\right)^4\right]^{-2}\)

- Use the property \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\):
\[
\left(\frac{2}{7}\right)^{-2} = \left(\frac{7}{2}\right)^2
\]
- Substitute this into the first part:
\[
\left[\left(\frac{2}{7}\right)^{-2}\right]^4 = \left[\left(\frac{7}{2}\right)^2\right]^4 = \left(\frac{7}{2}\right)^{2 \cdot 4} = \left(\frac{7}{2}\right)^8
\]
- For the second part:
\[
\left[\left(\frac{7}{2}\right)^4\right]^{-2} = \left(\frac{7}{2}\right)^{4 \cdot (-2)} = \left(\frac{7}{2}\right)^{-8}
\]
- Combine the two parts:
\[
\left[\left(\frac{2}{7}\right)^{-2}\right]^4 \times \left[\left(\frac{7}{2}\right)^4\right]^{-2} = \left(\frac{7}{2}\right)^8 \times \left(\frac{7}{2}\right)^{-8}
\]
- Use the property \(a^m \times a^n = a^{m+n}\):
\[
\left(\frac{7}{2}\right)^8 \times \left(\frac{7}{2}\right)^{-8} = \left(\frac{7}{2}\right)^{8 + (-8)} = \left(\frac{7}{2}\right)^0 = 1
\]

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8. If \(3^x = 243\), then find the value of \(x\).

- Express 243 as a power of 3:
\[
243 = 3^5
\]
- Therefore:
\[
3^x = 3^5
\]
- Since the bases are the same, equate the exponents:
\[
x = 5
\]

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Final Answers:

1. (a) \(81\), (b) \(26\), (c) \(128\), (d) \(121\), (e) \(2401\)
2. \(1 \times 10^3 + 7 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 + 5 \times 10^{-1} + 6 \times 10^{-2}\)
3. \(32\)
4. (a) \(3^{-3}\), (b) \(9\), (c) \(5^{-7}\), (d) \(676\)
5. \(\frac{1}{3}\)
6. (a) False, (b) False, (c) False, (d) True
7. \(1\)
8. \(5\)

\[
\boxed{5}
\]