Math worksheet on power and exponent concepts, including evaluation, simplification, and exponential form exercises.

Problem: Solve the given problems involving powers and exponents.

#### Section 1: Evaluate the following

1. (51) \(-(3)^{-2}\)
- Recall that \(a^{-n} = \frac{1}{a^n}\).
- Therefore, \((3)^{-2} = \frac{1}{3^2} = \frac{1}{9}\).
- So, \(-(3)^{-2} = -\frac{1}{9}\).

Answer: \(\boxed{-\frac{1}{9}}\)

2. (53) \((-7)^{-2}\)
- Using the rule \(a^{-n} = \frac{1}{a^n}\):
\[
(-7)^{-2} = \frac{1}{(-7)^2} = \frac{1}{49}
\]

Answer: \(\boxed{\frac{1}{49}}\)

3. (54) \((-11)^{-2}\)
- Similarly:
\[
(-11)^{-2} = \frac{1}{(-11)^2} = \frac{1}{121}
\]

Answer: \(\boxed{\frac{1}{121}}\)

4. (55) \(\left(\frac{1}{3}\right)^{-4}\)
- Using the rule \(\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n\):
\[
\left(\frac{1}{3}\right)^{-4} = \left(\frac{3}{1}\right)^4 = 3^4 = 81
\]

Answer: \(\boxed{81}\)

5. (56) \(\left(\frac{1}{2}\right)^{-4}\)
- Similarly:
\[
\left(\frac{1}{2}\right)^{-4} = \left(\frac{2}{1}\right)^4 = 2^4 = 16
\]

Answer: \(\boxed{16}\)

6. (57) \(\left(\frac{1}{1}\right)^{-4}\)
- Simplify the fraction first:
\[
\left(\frac{1}{1}\right)^{-4} = 1^{-4} = 1
\]

Answer: \(\boxed{1}\)

7. (58) \(\left(\frac{0}{3}\right)^{-4}\)
- The fraction \(\frac{0}{3} = 0\). Any non-zero number raised to a negative power is undefined if the base is zero.
- Therefore, \(\left(\frac{0}{3}\right)^{-4}\) is undefined.

Answer: \(\boxed{\text{undefined}}\)

8. (59) \(\left(\frac{-1}{2}\right)^{-1}\)
- Using the rule \(\left(\frac{a}{b}\right)^{-1} = \frac{b}{a}\):
\[
\left(\frac{-1}{2}\right)^{-1} = \frac{2}{-1} = -2
\]

Answer: \(\boxed{-2}\)

9. (60) \(\left(\frac{-1}{5}\right)^{-1}\)
- Similarly:
\[
\left(\frac{-1}{5}\right)^{-1} = \frac{5}{-1} = -5
\]

Answer: \(\boxed{-5}\)

10. (61) \(\left(\frac{-1}{7}\right)^{-1}\)
- Similarly:
\[
\left(\frac{-1}{7}\right)^{-1} = \frac{7}{-1} = -7
\]

Answer: \(\boxed{-7}\)

---

#### Section 2: Long question: Find the values of the following

11. (62) \(3^{-1} + 4^{-1}\)
- Using \(a^{-1} = \frac{1}{a}\):
\[
3^{-1} = \frac{1}{3}, \quad 4^{-1} = \frac{1}{4}
\]
Add the fractions:
\[
3^{-1} + 4^{-1} = \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12}
\]

Answer: \(\boxed{\frac{7}{12}}\)

12. (63) \((3^0 + 4^{-1}) \times 2^2\)
- Recall that \(a^0 = 1\):
\[
3^0 = 1, \quad 4^{-1} = \frac{1}{4}
\]
Add the terms inside the parentheses:
\[
3^0 + 4^{-1} = 1 + \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{5}{4}
\]
Multiply by \(2^2\):
\[
(3^0 + 4^{-1}) \times 2^2 = \frac{5}{4} \times 4 = 5
\]

Answer: \(\boxed{5}\)

13. (64) \((3^{-1} + 4^{-1} + 5^{-1})^0\)
- Any non-zero number raised to the power of 0 is 1:
\[
(3^{-1} + 4^{-1} + 5^{-1})^0 = 1
\]

Answer: \(\boxed{1}\)

14. (65) \(\left\{\left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\}^{-1}\)
- Simplify each term inside the braces:
\[
\left(\frac{1}{3}\right)^{-1} = 3, \quad \left(\frac{1}{4}\right)^{-1} = 4
\]
Subtract the terms:
\[
\left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1} = 3 - 4 = -1
\]
Take the reciprocal:
\[
\left\{\left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\}^{-1} = (-1)^{-1} = -1
\]

Answer: \(\boxed{-1}\)

---

#### Section 3: Simplify the following

15. (66) \(\left(4^{-1} \times 3^{-1}\right)^2\)
- Simplify inside the parentheses:
\[
4^{-1} = \frac{1}{4}, \quad 3^{-1} = \frac{1}{3}
\]
Multiply the fractions:
\[
4^{-1} \times 3^{-1} = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}
\]
Raise to the power of 2:
\[
\left(4^{-1} \times 3^{-1}\right)^2 = \left(\frac{1}{12}\right)^2 = \frac{1}{144}
\]

Answer: \(\boxed{\frac{1}{144}}\)

16. (67) \(\left(5^{-1} \div 6^{-1}\right)^3\)
- Simplify inside the parentheses:
\[
5^{-1} = \frac{1}{5}, \quad 6^{-1} = \frac{1}{6}
\]
Divide the fractions:
\[
5^{-1} \div 6^{-1} = \frac{1}{5} \div \frac{1}{6} = \frac{1}{5} \times \frac{6}{1} = \frac{6}{5}
\]
Raise to the power of 3:
\[
\left(5^{-1} \div 6^{-1}\right)^3 = \left(\frac{6}{5}\right)^3 = \frac{6^3}{5^3} = \frac{216}{125}
\]

Answer: \(\boxed{\frac{216}{125}}\)

17. (68) \(\left(2^{-1} + 3^{-1}\right)^{-1}\)
- Simplify inside the parentheses:
\[
2^{-1} = \frac{1}{2}, \quad 3^{-1} = \frac{1}{3}
\]
Add the fractions:
\[
2^{-1} + 3^{-1} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}
\]
Take the reciprocal:
\[
\left(2^{-1} + 3^{-1}\right)^{-1} = \left(\frac{5}{6}\right)^{-1} = \frac{6}{5}
\]

Answer: \(\boxed{\frac{6}{5}}\)

18. (69) \(\left(3^{-1} \times 4^{-1}\right)^{-1} \times 5^{-1}\)
- Simplify inside the parentheses:
\[
3^{-1} = \frac{1}{3}, \quad 4^{-1} = \frac{1}{4}
\]
Multiply the fractions:
\[
3^{-1} \times 4^{-1} = \frac{1}{3} \times \frac{1}{4} = \frac{1}{12}
\]
Take the reciprocal:
\[
\left(3^{-1} \times 4^{-1}\right)^{-1} = \left(\frac{1}{12}\right)^{-1} = 12
\]
Multiply by \(5^{-1}\):
\[
\left(3^{-1} \times 4^{-1}\right)^{-1} \times 5^{-1} = 12 \times \frac{1}{5} = \frac{12}{5}
\]

Answer: \(\boxed{\frac{12}{5}}\)

19. (70) \(\left(3^2 + 2^2\right) \times \left(\frac{1}{2}\right)^3\)
- Simplify inside the parentheses:
\[
3^2 = 9, \quad 2^2 = 4
\]
Add the terms:
\[
3^2 + 2^2 = 9 + 4 = 13
\]
Simplify the second term:
\[
\left(\frac{1}{2}\right)^3 = \frac{1}{2^3} = \frac{1}{8}
\]
Multiply the results:
\[
\left(3^2 + 2^2\right) \times \left(\frac{1}{2}\right)^3 = 13 \times \frac{1}{8} = \frac{13}{8}
\]

Answer: \(\boxed{\frac{13}{8}}\)

---

#### Section 4: Write the following in exponential form

20. (74) \(\left(\frac{3}{2}\right)^{-1} \times \left(\frac{3}{2}\right)^{-1} \times \left(\frac{3}{2}\right)^{-1} \times \left(\frac{3}{2}\right)^{-1}\)
- Use the property of exponents \(a^m \times a^n = a^{m+n}\):
\[
\left(\frac{3}{2}\right)^{-1} \times \left(\frac{3}{2}\right)^{-1} \times \left(\frac{3}{2}\right)^{-1} \times \left(\frac{3}{2}\right)^{-1} = \left(\frac{3}{2}\right)^{-1 + (-1) + (-1) + (-1)} = \left(\frac{3}{2}\right)^{-4}
\]

Answer: \(\boxed{\left(\frac{3}{2}\right)^{-4}}\)

21. (75) \(\left(\frac{2}{5}\right)^{-2} \times \left(\frac{2}{5}\right)^{-2} \times \left(\frac{2}{5}\right)^{-2}\)
- Use the same property of exponents:
\[
\left(\frac{2}{5}\right)^{-2} \times \left(\frac{2}{5}\right)^{-2} \times \left(\frac{2}{5}\right)^{-2} = \left(\frac{2}{5}\right)^{-2 + (-2) + (-2)} = \left(\frac{2}{5}\right)^{-6}
\]

Answer: \(\boxed{\left(\frac{2}{5}\right)^{-6}}\)

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{(51)} -\frac{1}{9}, \quad \text{(53)} \frac{1}{49}, \quad \text{(54)} \frac{1}{121}, \quad \text{(55)} 81, \quad \text{(56)} 16, \quad \text{(57)} 1, \quad \text{(58)} \text{undefined}, \\
&\text{(59)} -2, \quad \text{(60)} -5, \quad \text{(61)} -7, \quad \text{(62)} \frac{7}{12}, \quad \text{(63)} 5, \quad \text{(64)} 1, \quad \text{(65)} -1, \\
&\text{(66)} \frac{1}{144}, \quad \text{(67)} \frac{216}{125}, \quad \text{(68)} \frac{6}{5}, \quad \text{(69)} \frac{12}{5}, \quad \text{(70)} \frac{13}{8}, \\
&\text{(74)} \left(\frac{3}{2}\right)^{-4}, \quad \text{(75)} \left(\frac{2}{5}\right)^{-6}.
\end{aligned}
}
\]