Worksheet on simplifying and evaluating exponential expressions.

Problem: Simplify and solve the given expressions involving exponents.

#### Section 1: Simplify and write in exponential form

We will simplify each expression step by step using the rules of exponents:
1. $ a^m \times a^n = a^{m+n} $
2. $ a^m \div a^n = a^{m-n} $
3. $ (a^m)^n = a^{m \cdot n} $
4. $ a^{-n} = \frac{1}{a^n} $
5. $ a^0 = 1 $ (for any $ a \neq 0 $)

---

#### Part 1(a): Simplify $ (-5)^3 \times (-5)^4 $

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

Answer: $ (-5)^7 $

---

#### Part 1(b): Simplify $ P^3 \times P^{-10} $

Using the rule $ a^m \times a^n = a^{m+n} $:
$$
P^3 \times P^{-10} = P^{3 + (-10)} = P^{-7}
$$

Using the rule $ a^{-n} = \frac{1}{a^n} $:
$$
P^{-7} = \frac{1}{P^7}
$$

Answer: $ \frac{1}{P^7} $

---

#### Part 1(c): Simplify $ 3^5 \times 3^{-10} \times 3^6 $

Using the rule $ a^m \times a^n = a^{m+n} $:
$$
3^5 \times 3^{-10} \times 3^6 = 3^{5 + (-10) + 6} = 3^{5 - 10 + 6} = 3^1 = 3
$$

Answer: $ 3 $

---

#### Part 1(d): Simplify $ (2^5 \div 2^8)^5 \times 2^{-5} $

First, simplify $ 2^5 \div 2^8 $ using the rule $ a^m \div a^n = a^{m-n} $:
$$
2^5 \div 2^8 = 2^{5-8} = 2^{-3}
$$

Now raise it to the power of 5:
$$
(2^{-3})^5 = 2^{-3 \cdot 5} = 2^{-15}
$$

Next, multiply by $ 2^{-5} $:
$$
2^{-15} \times 2^{-5} = 2^{-15 + (-5)} = 2^{-20}
$$

Using the rule $ a^{-n} = \frac{1}{a^n} $:
$$
2^{-20} = \frac{1}{2^{20}}
$$

Answer: $ \frac{1}{2^{20}} $

---

#### Part 1(e): Simplify $ (-4)^{-3} \times (5)^3 \times (-5)^{-3} $

First, handle each term separately:
- $ (-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64} $
- $ (5)^3 = 125 $
- $ (-5)^{-3} = \frac{1}{(-5)^3} = \frac{1}{-125} = -\frac{1}{125} $

Now multiply these together:
$$
(-4)^{-3} \times (5)^3 \times (-5)^{-3} = \left(-\frac{1}{64}\right) \times 125 \times \left(-\frac{1}{125}\right)
$$

Simplify step by step:
$$
\left(-\frac{1}{64}\right) \times 125 = -\frac{125}{64}
$$
$$
-\frac{125}{64} \times \left(-\frac{1}{125}\right) = \frac{125}{64} \times \frac{1}{125} = \frac{1}{64}
$$

Answer: $ \frac{1}{64} $

---

#### Part 1(f): Simplify $ (-3)^4 \times \left(\frac{2}{3}\right)^4 $

Using the rule $ (ab)^n = a^n \times b^n $:
$$
(-3)^4 \times \left(\frac{2}{3}\right)^4 = \left((-3) \times \frac{2}{3}\right)^4
$$

Simplify inside the parentheses:
$$
(-3) \times \frac{2}{3} = -2
$$

So:
$$
\left((-3) \times \frac{2}{3}\right)^4 = (-2)^4
$$

Calculate $ (-2)^4 $:
$$
(-2)^4 = 16
$$

Answer: $ 16 $

---

#### Part 1(g): Simplify $ \frac{1}{8} \times 3^{-5} $

Rewrite $ \frac{1}{8} $ as $ 2^{-3} $:
$$
\frac{1}{8} = 2^{-3}
$$

So the expression becomes:
$$
2^{-3} \times 3^{-5}
$$

Since the bases are different, we cannot combine the exponents. The expression is already in its simplest form:
$$
2^{-3} \times 3^{-5}
$$

Using the rule $ a^{-n} = \frac{1}{a^n} $:
$$
2^{-3} \times 3^{-5} = \frac{1}{2^3} \times \frac{1}{3^5} = \frac{1}{8} \times \frac{1}{243} = \frac{1}{1944}
$$

Answer: $ \frac{1}{1944} $

---

#### Part 1(h): Simplify $ (-4)^5 \div (4)^8 $

First, note that $ (-4)^5 = -(4^5) $ because the exponent is odd:
$$
(-4)^5 = -(4^5)
$$

So the expression becomes:
$$
\frac{-(4^5)}{4^8}
$$

Using the rule $ a^m \div a^n = a^{m-n} $:
$$
\frac{4^5}{4^8} = 4^{5-8} = 4^{-3}
$$

Thus:
$$
\frac{-(4^5)}{4^8} = -4^{-3}
$$

Using the rule $ a^{-n} = \frac{1}{a^n} $:
$$
-4^{-3} = -\frac{1}{4^3} = -\frac{1}{64}
$$

Answer: $ -\frac{1}{64} $

---

Section 2: Find the value of

#### Part 2(a): Simplify $ \left(\frac{2}{3}\right)^{-2} $

Using the rule $ \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n $:
$$
\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}
$$

Answer: $ \frac{9}{4} $

---

#### Part 2(b): Simplify $ \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} = \left(\frac{3}{1}\right)^2 = 3^2 = 9 $
- $ \left(\frac{1}{2}\right)^{-3} = \left(\frac{2}{1}\right)^3 = 2^3 = 8 $
- $ \left(\frac{1}{4}\right)^{-2} = \left(\frac{4}{1}\right)^2 = 4^2 = 16 $

Now substitute back:
$$
\left[\left(\frac{1}{3}\right)^{-2} - \left(\frac{1}{2}\right)^{-3}\right] \div \left(\frac{1}{4}\right)^{-2} = [9 - 8] \div 16 = 1 \div 16 = \frac{1}{16}
$$

Answer: $ \frac{1}{16} $

---

#### Part 2(c): Simplify $ \left(\frac{5}{8}\right)^{-7} \times \left(\frac{8}{5}\right)^{-5} $

Using the rule $ \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n $:
- $ \left(\frac{5}{8}\right)^{-7} = \left(\frac{8}{5}\right)^7 $
- $ \left(\frac{8}{5}\right)^{-5} = \left(\frac{5}{8}\right)^5 $

So the expression becomes:
$$
\left(\frac{8}{5}\right)^7 \times \left(\frac{5}{8}\right)^5
$$

Using the rule $ a^m \times a^n = a^{m+n} $ for the same base:
$$
\left(\frac{8}{5}\right)^7 \times \left(\frac{5}{8}\right)^5 = \left(\frac{8}{5}\right)^{7-5} = \left(\frac{8}{5}\right)^2 = \frac{8^2}{5^2} = \frac{64}{25}
$$

Answer: $ \frac{64}{25} $

---

#### Part 2(d): Simplify $ (3^0 + 4^{-1}) \times 2^2 $

First, simplify each term:
- $ 3^0 = 1 $
- $ 4^{-1} = \frac{1}{4} $
- $ 2^2 = 4 $

So the expression becomes:
$$
(3^0 + 4^{-1}) \times 2^2 = \left(1 + \frac{1}{4}\right) \times 4
$$

Simplify $ 1 + \frac{1}{4} $:
$$
1 + \frac{1}{4} = \frac{4}{4} + \frac{1}{4} = \frac{5}{4}
$$

Now multiply:
$$
\left(\frac{5}{4}\right) \times 4 = \frac{5}{4} \times \frac{4}{1} = 5
$$

Answer: $ 5 $

---

#### Part 2(e): Simplify $ (2^{-1} \times 4^{-1}) \div 2^{-2} $

First, simplify $ 2^{-1} \times 4^{-1} $:
- $ 4^{-1} = (2^2)^{-1} = 2^{-2} $

So:
$$
2^{-1} \times 4^{-1} = 2^{-1} \times 2^{-2} = 2^{-1-2} = 2^{-3}
$$

Now divide by $ 2^{-2} $:
$$
\frac{2^{-3}}{2^{-2}} = 2^{-3 - (-2)} = 2^{-3 + 2} = 2^{-1}
$$

Using the rule $ a^{-n} = \frac{1}{a^n} $:
$$
2^{-1} = \frac{1}{2}
$$

Answer: $ \frac{1}{2} $

---

#### Part 2(f): Simplify $ \left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} $

Using the rule $ \left(\frac{1}{a}\right)^{-n} = a^n $:
- $ \left(\frac{1}{2}\right)^{-2} = 2^2 = 4 $
- $ \left(\frac{1}{3}\right)^{-2} = 3^2 = 9 $
- $ \left(\frac{1}{4}\right)^{-2} = 4^2 = 16 $

So the expression becomes:
$$
4 + 9 + 16 = 29
$$

Answer: $ 29 $

---

#### Part 2(g): Simplify $ (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: $ 1 $

---

#### Part 2(h): Simplify $ \left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2 $

First, simplify $ \left(\frac{-2}{3}\right)^{-2} $:
$$
\left(\frac{-2}{3}\right)^{-2} = \left(\frac{3}{-2}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{3^2}{2^2} = \frac{9}{4}
$$

Now raise it to the power of 2:
$$
\left\{\left(\frac{-2}{3}\right)^{-2}\right\}^2 = \left(\frac{9}{4}\right)^2 = \frac{9^2}{4^2} = \frac{81}{16}
$$

Answer: $ \frac{81}{16} $

---

#### Part 2(i): Simplify $ \frac{8^{-1} \times 5^3}{2^{-4}} $

First, simplify each term:
- $ 8^{-1} = \frac{1}{8} $
- $ 5^3 = 125 $
- $ 2^{-4} = \frac{1}{2^4} = \frac{1}{16} $

So the expression becomes:
$$
\frac{8^{-1} \times 5^3}{2^{-4}} = \frac{\frac{1}{8} \times 125}{\frac{1}{16}}
$$

Simplify the numerator:
$$
\frac{1}{8} \times 125 = \frac{125}{8}
$$

Now divide by $ \frac{1}{16} $:
$$
\frac{\frac{125}{8}}{\frac{1}{16}} = \frac{125}{8} \times \frac{16}{1} = \frac{125 \times 16}{8} = \frac{2000}{8} = 250
$$

Answer: $ 250 $

---

#### Part 2(j): Simplify $ (5^{-1} \times 2^{-1}) \times 6^{-1} $

First, simplify each term:
- $ 5^{-1} = \frac{1}{5} $
- $ 2^{-1} = \frac{1}{2} $
- $ 6^{-1} = \frac{1}{6} $

So the expression becomes:
$$
(5^{-1} \times 2^{-1}) \times 6^{-1} = \left(\frac{1}{5} \times \frac{1}{2}\right) \times \frac{1}{6}
$$

Simplify step by step:
$$
\frac{1}{5} \times \frac{1}{2} = \frac{1}{10}
$$
$$
\frac{1}{10} \times \frac{1}{6} = \frac{1}{60}
$$

Answer: $ \frac{1}{60} $

---

#### Part 2(k): Simplify $ \left\{\left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\}^{-1} $

First, simplify each term:
- $ \left(\frac{1}{3}\right)^{-1} = 3 $
- $ \left(\frac{1}{4}\right)^{-1} = 4 $

So the expression becomes:
$$
\left\{\left(\frac{1}{3}\right)^{-1} - \left(\frac{1}{4}\right)^{-1}\right\}^{-1} = \{3 - 4\}^{-1} = (-1)^{-1}
$$

Using the rule $ a^{-1} = \frac{1}{a} $:
$$
(-1)^{-1} = \frac{1}{-1} = -1
$$

Answer: $ -1 $

---

#### Part 2(l): Simplify $ \left(\frac{5}{8}\right)^{-7} \times \left(\frac{8}{5}\right)^{-4} $

Using the rule $ \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n $:
- $ \left(\frac{5}{8}\right)^{-7} = \left(\frac{8}{5}\right)^7 $
- $ \left(\frac{8}{5}\right)^{-4} = \left(\frac{5}{8}\right)^4 $

So the expression becomes:
$$
\left(\frac{8}{5}\right)^7 \times \left(\frac{5}{8}\right)^4
$$

Using the rule $ a^m \times a^n = a^{m+n} $ for the same base:
$$
\left(\frac{8}{5}\right)^7 \times \left(\frac{5}{8}\right)^4 = \left(\frac{8}{5}\right)^{7-4} = \left(\frac{8}{5}\right)^3 = \frac{8^3}{5^3} = \frac{512}{125}
$$

Answer: $ \frac{512}{125} $

---

#### Part 2(m): Simplify $ 25 \times t^{-4} $

Rewrite $ 25 $ as $ 5^2 $:
$$
25 \times t^{-4} = 5^2 \times t^{-4}
$$

This is already in its simplest form:
$$
5^2 \times t^{-4} = \frac{25}{t^4}
$$

Answer: $ \frac{25}{t^4} $

---

#### Part 2(n): Simplify $ \frac{3^{-5} \times 10^{-5} \times 125}{1} $

First, simplify each term:
- $ 3^{-5} = \frac{1}{3^5} = \frac{1}{243} $
- $ 10^{-5} = \frac{1}{10^5} = \frac{1}{100000} $
- $ 125 = 5^3 $

So the expression becomes:
$$
\frac{3^{-5} \times 10^{-5} \times 125}{1} = \frac{\frac{1}{243} \times \frac{1}{100000} \times 5^3}{1}
$$

Simplify step by step:
$$
\frac{1}{243} \times \frac{1}{100000} = \frac{1}{243 \times 100000} = \frac{1}{24300000}
$$
$$
\frac{1}{24300000} \times 5^3 = \frac{5^3}{24300000} = \frac{125}{24300000}
$$

Simplify the fraction:
$$
\frac{125}{24300000} = \frac{1}{194400}
$$

Answer: $ \frac{1}{194400} $

---

Final Answers:

1. (a) $ (-5)^7 $
(b) $ \frac{1}{P^7} $
(c) $ 3 $
(d) $ \frac{1}{2^{20}} $
(e) $ \frac{1}{64} $
(f) $ 16 $
(g) $ \frac{1}{1944} $
(h) $ -\frac{1}{64} $

2. (a) $ \frac{9}{4} $
(b) $ \frac{1}{16} $
(c) $ \frac{64}{25} $
(d) $ 5 $
(e) $ \frac{1}{2} $
(f) $ 29 $
(g) $ 1 $
(h) $ \frac{81}{16} $
(i) $ 250 $
(j) $ \frac{1}{60} $
(k) $ -1 $
(l) $ \frac{512}{125} $
(m) $ \frac{25}{t^4} $
(n) $ \frac{1}{194400} $

Boxed Final Answer:
$$
\boxed{
\begin{aligned}
&\text{1. (a) } (-5)^7, \quad \text{(b) } \frac{1}{P^7}, \quad \text{(c) } 3, \quad \text{(d) } \frac{1}{2^{20}}, \quad \text{(e) } \frac{1}{64}, \quad \text{(f) } 16, \quad \text{(g) } \frac{1}{1944}, \quad \text{(h) } -\frac{1}{64} \\
&\text{2. (a) } \frac{9}{4}, \quad \text{(b) } \frac{1}{16}, \quad \text{(c) } \frac{64}{25}, \quad \text{(d) } 5, \quad \text{(e) } \frac{1}{2}, \quad \text{(f) } 29, \quad \text{(g) } 1, \quad \text{(h) } \frac{81}{16}, \\
&\quad \text{(i) } 250, \quad \text{(j) } \frac{1}{60}, \quad \text{(k) } -1, \quad \text{(l) } \frac{512}{125}, \quad \text{(m) } \frac{25}{t^4}, \quad \text{(n) } \frac{1}{194400}
\end{aligned}
}
$$