Comprehensive math worksheet covering various operations with exponents, including simplification and evaluation.

Problem Set: Exponents (Powers)

Below, I will solve each problem step by step, explaining the reasoning and applying the rules of exponents.

---

1. Simplify and write in exponential form:

#### a. \( (-5)^3 \times (-5)^4 \)
Using the rule \( a^m \times a^n = a^{m+n} \):
\[
(-5)^3 \times (-5)^4 = (-5)^{3+4} = (-5)^7
\]

#### b. \( P^3 \times P^{-10} \)
Using the same rule:
\[
P^3 \times P^{-10} = P^{3 + (-10)} = P^{-7}
\]

#### c. \( 3^5 \times 3^{-10} \times 3^6 \)
Combine the exponents:
\[
3^5 \times 3^{-10} \times 3^6 = 3^{5 + (-10) + 6} = 3^{1}
\]

#### d. \( (2^5 \div 2^8) \times 2^{-5} \)
First, simplify \( 2^5 \div 2^8 \) using \( a^m \div a^n = a^{m-n} \):
\[
2^5 \div 2^8 = 2^{5-8} = 2^{-3}
\]
Now multiply:
\[
2^{-3} \times 2^{-5} = 2^{-3 + (-5)} = 2^{-8}
\]

#### e. \( (-4)^{-3} \times (5)^3 \times (-5)^{-3} \)
Simplify each term:
\[
(-4)^{-3} = \frac{1}{(-4)^3}, \quad (5)^3 = 5^3, \quad (-5)^{-3} = \frac{1}{(-5)^3}
\]
Combine:
\[
(-4)^{-3} \times (5)^3 \times (-5)^{-3} = \frac{1}{(-4)^3} \times 5^3 \times \frac{1}{(-5)^3}
\]
\[
= \frac{5^3}{(-4)^3 \times (-5)^3} = \frac{5^3}{(-1)^3 \times 4^3 \times 5^3} = \frac{5^3}{-1 \times 4^3 \times 5^3} = \frac{1}{-4^3} = -\frac{1}{64}
\]

#### f. \( (-3)^4 \times \left(\frac{2}{3}\right)^4 \)
Use the property \( (ab)^n = a^n \times b^n \):
\[
(-3)^4 \times \left(\frac{2}{3}\right)^4 = \left((-3) \times \frac{2}{3}\right)^4 = \left(-2\right)^4 = 16
\]

#### g. \( \frac{1}{8} \times 3^{-5} \)
Rewrite \( \frac{1}{8} \) as \( 2^{-3} \):
\[
\frac{1}{8} \times 3^{-5} = 2^{-3} \times 3^{-5}
\]
This is already in exponential form:
\[
2^{-3} \times 3^{-5}
\]

#### h. \( (-4)^5 \div (4)^8 \)
Simplify the division:
\[
(-4)^5 \div (4)^8 = \frac{(-4)^5}{4^8}
\]
Since \( 4 = 2^2 \), rewrite:
\[
(-4)^5 = (-2^2)^5 = (-1)^5 \times (2^2)^5 = -2^{10}, \quad 4^8 = (2^2)^8 = 2^{16}
\]
So:
\[
\frac{(-4)^5}{4^8} = \frac{-2^{10}}{2^{16}} = -2^{10-16} = -2^{-6}
\]

---

2. Find the value of:

#### a. \( \left(\frac{2}{3}\right)^{-2} \)
Using \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \):
\[
\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}
\]

#### b. \( \left[\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2} \)
First, simplify each term:
\[
\left(\frac{1}{3}\right)^{-2} = 3^2 = 9, \quad \left(\frac{1}{2}\right)^{-3} = 2^3 = 8, \quad \left(\frac{1}{4}\right)^{-2} = 4^2 = 16
\]
Now substitute:
\[
\left[9 - 8\right] \div 16 = 1 \div 16 = \frac{1}{16}
\]

#### c. \( \left(\frac{5}{8}\right)^{-7} \times \left(\frac{8}{5}\right)^{-5} \)
Using \( \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n \):
\[
\left(\frac{5}{8}\right)^{-7} = \left(\frac{8}{5}\right)^7, \quad \left(\frac{8}{5}\right)^{-5} = \left(\frac{5}{8}\right)^5
\]
So:
\[
\left(\frac{8}{5}\right)^7 \times \left(\frac{5}{8}\right)^5 = \left(\frac{8}{5} \times \frac{5}{8}\right)^5 \times \left(\frac{8}{5}\right)^2 = 1^5 \times \left(\frac{8}{5}\right)^2 = \left(\frac{8}{5}\right)^2 = \frac{64}{25}
\]

#### d. \( (3^0 + 4^{-1}) \times 2^2 \)
Simplify each term:
\[
3^0 = 1, \quad 4^{-1} = \frac{1}{4}, \quad 2^2 = 4
\]
So:
\[
(3^0 + 4^{-1}) \times 2^2 = \left(1 + \frac{1}{4}\right) \times 4 = \left(\frac{4}{4} + \frac{1}{4}\right) \times 4 = \frac{5}{4} \times 4 = 5
\]

#### e. \( (2^{-1} \times 4^{-1}) \div 2^{-2} \)
Simplify each term:
\[
2^{-1} = \frac{1}{2}, \quad 4^{-1} = \frac{1}{4} = \frac{1}{2^2} = 2^{-2}, \quad 2^{-2} = \frac{1}{4}
\]
So:
\[
2^{-1} \times 4^{-1} = 2^{-1} \times 2^{-2} = 2^{-1-2} = 2^{-3} = \frac{1}{8}
\]
Now divide:
\[
\frac{1}{8} \div 2^{-2} = \frac{1}{8} \times 2^2 = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2}
\]

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

#### g. \( \frac{8^{-1} \times 5^3}{2^{-4}} \)
Simplify each term:
\[
8^{-1} = \frac{1}{8}, \quad 5^3 = 125, \quad 2^{-4} = \frac{1}{16}
\]
So:
\[
\frac{8^{-1} \times 5^3}{2^{-4}} = \frac{\frac{1}{8} \times 125}{\frac{1}{16}} = \frac{125}{8} \times 16 = \frac{125 \times 16}{8} = \frac{2000}{8} = 250
\]

#### h. \( \left\{ \left( \frac{-2}{3} \right)^{-2} \right\}^2 \)
First, simplify \( \left( \frac{-2}{3} \right)^{-2} \):
\[
\left( \frac{-2}{3} \right)^{-2} = \left( \frac{3}{-2} \right)^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4}
\]
Now square:
\[
\left( \frac{9}{4} \right)^2 = \frac{81}{16}
\]

#### i. \( \left( \frac{1}{3} \right)^{-1} - \left( \frac{1}{4} \right)^{-1} \)
Simplify each term:
\[
\left( \frac{1}{3} \right)^{-1} = 3, \quad \left( \frac{1}{4} \right)^{-1} = 4
\]
So:
\[
3 - 4 = -1
\]

#### j. \( (5^{-1} \times 2^{-1}) \times 6^{-1} \)
Simplify each term:
\[
5^{-1} = \frac{1}{5}, \quad 2^{-1} = \frac{1}{2}, \quad 6^{-1} = \frac{1}{6}
\]
So:
\[
(5^{-1} \times 2^{-1}) \times 6^{-1} = \left( \frac{1}{5} \times \frac{1}{2} \right) \times \frac{1}{6} = \frac{1}{10} \times \frac{1}{6} = \frac{1}{60}
\]

#### k. \( \left( \frac{5}{8} \right)^{-7} \times \left( \frac{8}{5} \right)^{-4} \)
Using \( \left( \frac{a}{b} \right)^{-n} = \left( \frac{b}{a} \right)^n \):
\[
\left( \frac{5}{8} \right)^{-7} = \left( \frac{8}{5} \right)^7, \quad \left( \frac{8}{5} \right)^{-4} = \left( \frac{5}{8} \right)^4
\]
So:
\[
\left( \frac{8}{5} \right)^7 \times \left( \frac{5}{8} \right)^4 = \left( \frac{8}{5} \times \frac{5}{8} \right)^4 \times \left( \frac{8}{5} \right)^3 = 1^4 \times \left( \frac{8}{5} \right)^3 = \left( \frac{8}{5} \right)^3 = \frac{512}{125}
\]

#### l. \( \frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} \)
Simplify each term:
\[
25 = 5^2, \quad 5^{-3} = \frac{1}{5^3}, \quad 10 = 2 \times 5
\]
So:
\[
\frac{25 \times t^{-4}}{5^{-3} \times 10 \times t^{-8}} = \frac{5^2 \times t^{-4}}{\frac{1}{5^3} \times 2 \times 5 \times t^{-8}} = \frac{5^2 \times t^{-4}}{\frac{2 \times 5}{5^3} \times t^{-8}} = \frac{5^2 \times t^{-4}}{\frac{2}{5^2} \times t^{-8}}
\]
\[
= \frac{5^2 \times t^{-4} \times 5^2}{2 \times t^{-8}} = \frac{5^4 \times t^{-4+8}}{2} = \frac{5^4 \times t^4}{2} = \frac{625t^4}{2}
\]

#### m. \( \frac{3^{-5} \times 10^{-5} \times 125}{6^{-5} \times 6^{-5}} \)
Simplify each term:
\[
3^{-5} = \frac{1}{3^5}, \quad 10^{-5} = \frac{1}{10^5}, \quad 125 = 5^3, \quad 6^{-5} = \frac{1}{6^5}
\]
So:
\[
\frac{3^{-5} \times 10^{-5} \times 125}{6^{-5} \times 6^{-5}} = \frac{\frac{1}{3^5} \times \frac{1}{10^5} \times 5^3}{\frac{1}{6^5} \times \frac{1}{6^5}} = \frac{5^3}{3^5 \times 10^5 \times \frac{1}{6^{10}}} = \frac{5^3 \times 6^{10}}{3^5 \times 10^5}
\]
Since \( 10^5 = (2 \times 5)^5 = 2^5 \times 5^5 \):
\[
= \frac{5^3 \times 6^{10}}{3^5 \times 2^5 \times 5^5} = \frac{6^{10}}{3^5 \times 2^5 \times 5^2} = \frac{(2 \times 3)^{10}}{3^5 \times 2^5 \times 5^2} = \frac{2^{10} \times 3^{10}}{3^5 \times 2^5 \times 5^2} = \frac{2^5 \times 3^5}{5^2} = \frac{32 \times 243}{25} = \frac{7776}{25}
\]

---

3. Evaluate:

#### a. \( 3^{-2} \)
\[
3^{-2} = \frac{1}{3^2} = \frac{1}{9}
\]

#### b. \( (-4)^{-2} \)
\[
(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}
\]

#### c. \( \left( \frac{1}{2} \right)^{-5} \)
\[
\left( \frac{1}{2} \right)^{-5} = 2^5 = 32
\]

#### d. \( 1 / 3^{-2} \)
\[
3^{-2} = \frac{1}{9}, \quad \frac{1}{3^{-2}} = \frac{1}{\frac{1}{9}} = 9
\]

#### e. \( 2^{-3} \)
\[
2^{-3} = \frac{1}{2^3} = \frac{1}{8}
\]

---

4. Find the multiplicative inverse of the following:

#### a. \( 2^{-4} \)
The multiplicative inverse of \( 2^{-4} \) is \( 2^4 \):
\[
2^4 = 16
\]

#### b. \( 10^{-5} \)
The multiplicative inverse of \( 10^{-5} \) is \( 10^5 \):
\[
10^5 = 100000
\]

#### c. \( 7^{-2} \)
The multiplicative inverse of \( 7^{-2} \) is \( 7^2 \):
\[
7^2 = 49
\]

#### d. \( 5^{-3} \)
The multiplicative inverse of \( 5^{-3} \) is \( 5^3 \):
\[
5^3 = 125
\]

#### e. \( 10^{-100} \)
The multiplicative inverse of \( 10^{-100} \) is \( 10^{100} \):
\[
10^{100}
\]

---

5. If \( \frac{m}{n} = \left( \frac{5}{7} \right)^4 \div \left( \frac{5}{7} \right)^0 \), find the value of \( \left( \frac{m}{n} \right)^2 \) if \( m = -3 \) and \( n = 4 \).

First, simplify \( \left( \frac{5}{7} \right)^4 \div \left( \frac{5}{7} \right)^0 \):
\[
\left( \frac{5}{7} \right)^0 = 1, \quad \left( \frac{5}{7} \right)^4 \div 1 = \left( \frac{5}{7} \right)^4
\]
So:
\[
\frac{m}{n} = \left( \frac{5}{7} \right)^4
\]
Given \( m = -3 \) and \( n = 4 \):
\[
\frac{m}{n} = \frac{-3}{4}
\]
Thus:
\[
\left( \frac{m}{n} \right)^2 = \left( \frac{-3}{4} \right)^2 = \frac{9}{16}
\]

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6. Find the value of \( x^{-2} \) if \( x = \left( \frac{-2}{5} \right)^{-3} \div \left( \frac{5}{6} \right)^0 \).

First, simplify \( \left( \frac{-2}{5} \right)^{-3} \div \left( \frac{5}{6} \right)^0 \):
\[
\left( \frac{5}{6} \right)^0 = 1, \quad \left( \frac{-2}{5} \right)^{-3} = \left( \frac{5}{-2} \right)^3 = \left( \frac{5}{-2} \right)^3 = \frac{5^3}{(-2)^3} = \frac{125}{-8} = -\frac{125}{8}
\]
So:
\[
x = -\frac{125}{8}
\]
Now find \( x^{-2} \):
\[
x^{-2} = \left( -\frac{125}{8} \right)^{-2} = \left( \frac{8}{-125} \right)^2 = \frac{8^2}{(-125)^2} = \frac{64}{15625}
\]

---

7. What should \( \left( \frac{7}{9} \right)^{-3} \) be divided so that the quotient becomes 9?

Let the divisor be \( x \). We need:
\[
\frac{\left( \frac{7}{9} \right)^{-3}}{x} = 9
\]
Simplify \( \left( \frac{7}{9} \right)^{-3} \):
\[
\left( \frac{7}{9} \right)^{-3} = \left( \frac{9}{7} \right)^3 = \frac{9^3}{7^3} = \frac{729}{343}
\]
So:
\[
\frac{\frac{729}{343}}{x} = 9 \implies \frac{729}{343x} = 9 \implies 729 = 9 \times 343x \implies x = \frac{729}{9 \times 343} = \frac{81}{343}
\]

---

8. What should \( \left( \frac{2}{5} \right)^4 \) be multiplied so that the product becomes 25?

Let the multiplier be \( y \). We need:
\[
\left( \frac{2}{5} \right)^4 \times y = 25
\]
Simplify \( \left( \frac{2}{5} \right)^4 \):
\[
\left( \frac{2}{5} \right)^4 = \frac{2^4}{5^4} = \frac{16}{625}
\]
So:
\[
\frac{16}{625} \times y = 25 \implies y = 25 \times \frac{625}{16} = \frac{15625}{16}
\]

---

9. Express each of the following rational numbers in exponential form:

#### a. \( \frac{1}{343} \)
\[
343 = 7^3, \quad \frac{1}{343} = 7^{-3}
\]

#### b. \( \frac{-25}{216} \)
\[
25 = 5^2, \quad 216 = 6^3 = (2 \times 3)^3 = 2^3 \times 3^3
\]
So:
\[
\frac{-25}{216} = \frac{-5^2}{2^3 \times 3^3}
\]

#### c. \( \frac{64}{27} \)
\[
64 = 4^3 = (2^2)^3 = 2^6, \quad 27 = 3^3
\]
So:
\[
\frac{64}{27} = \frac{2^6}{3^3}
\]

#### d. \( \frac{243000}{729000} \)
Simplify the fraction:
\[
\frac{243000}{729000} = \frac{243}{729} = \frac{1}{3}
\]
So:
\[
\frac{1}{3} = 3^{-1}
\]

---

10. Simplify and express the result as powers of 2:

#### a. \( \left( \left( \frac{1}{2} \right)^3 \right)^4 \div \left( \left( \frac{1}{2} \right)^4 \right)^3 \)
Simplify each term:
\[
\left( \left( \frac{1}{2} \right)^3 \right)^4 = \left( \frac{1}{2} \right)^{3 \times 4} = \left( \frac{1}{2} \right)^{12} = 2^{-12}
\]
\[
\left( \left( \frac{1}{2} \right)^4 \right)^3 = \left( \frac{1}{2} \right)^{4 \times 3} = \left( \frac{1}{2} \right)^{12} = 2^{-12}
\]
So:
\[
\frac{2^{-12}}{2^{-12}} = 2^{-12 - (-12)} = 2^0 = 1
\]

#### b. \( (3^0 + 5^0) \div (4^0 + 2^0) \)
Simplify each term:
\[
3^0 = 1, \quad 5^0 = 1, \quad 4^0 = 1, \quad 2^0 = 1
\]
So:
\[
(3^0 + 5^0) \div (4^0 + 2^0) = (1 + 1) \div (1 + 1) = 2 \div 2 = 1
\]

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & \text{a. } (-5)^7, \text{ b. } P^{-7}, \text{ c. } 3^1, \text{ d. } 2^{-8}, \text{ e. } -\frac{1}{64}, \text{ f. } 16, \text{ g. } 2^{-3} \times 3^{-5}, \text{ h. } -2^{-6} \\
2. & \text{a. } \frac{9}{4}, \text{ b. } \frac{1}{16}, \text{ c. } \frac{64}{25}, \text{ d. } 5, \text{ e. } \frac{1}{2}, \text{ f. } 1, \text{ g. } 250, \text{ h. } \frac{81}{16}, \text{ i. } -1, \text{ j. } \frac{1}{60}, \text{ k. } \frac{512}{125}, \text{ l. } \frac{625t^4}{2}, \text{ m. } \frac{7776}{25} \\
3. & \text{a. } \frac{1}{9}, \text{ b. } \frac{1}{16}, \text{ c. } 32, \text{ d. } 9, \text{ e. } \frac{1}{8} \\
4. & \text{a. } 16, \text{ b. } 100000, \text{ c. } 49, \text{ d. } 125, \text{ e. } 10^{100} \\
5. & \frac{9}{16} \\
6. & \frac{64}{15625} \\
7. & \frac{81}{343} \\
8. & \frac{15625}{16} \\
9. & \text{a. } 7^{-3}, \text{ b. } \frac{-5^2}{2^3 \times 3^3}, \text{ c. } \frac{2^6}{3^3}, \text{ d. } 3^{-1} \\
10. & \text{a. } 1, \text{ b. } 1 \\
\end{array}
}
\]