Let’s solve the given problems step by step.

Exercise 12.1

1. Evaluate:

(i) $3^{-2}$:
$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

(ii) $(-4)^{-2}$:
$(-4)^{-2} = \frac{1}{(-4)^2} = \frac{1}{16}$

(iii) $\left(\frac{1}{2}\right)^{-5}$:
$\left(\frac{1}{2}\right)^{-5} = \left(\frac{2}{1}\right)^5 = 2^5 = 32$

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

(i) $(-4)^5 \div (-4)^8$:
Using the property $a^m \div a^n = a^{m-n}$:
$(-4)^5 \div (-4)^8 = (-4)^{5-8} = (-4)^{-3}$
Expressed as a positive exponent:
$(-4)^{-3} = \frac{1}{(-4)^3} = \frac{1}{-64}$

(ii) $\left(\frac{1}{2^3}\right)^2$:
Using the property $\left(\frac{a}{b}\right)^m = \frac{a^m}{b^m}$:
$\left(\frac{1}{2^3}\right)^2 = \frac{1^2}{(2^3)^2} = \frac{1}{2^6} = \frac{1}{64}$

(iii) $(-3)^4 \times \left(\frac{5}{3}\right)^4$:
$(-3)^4 \times \left(\frac{5}{3}\right)^4 = \frac{(-3)^4 \cdot 5^4}{3^4}$
Since $(-3)^4 = 3^4$:
$= \frac{3^4 \cdot 5^4}{3^4} = 5^4 = 625$

(iv) $(3^{-7} \div 3^{-10}) \times 3^{-5}$:
Using $a^m \div a^n = a^{m-n}$:
$3^{-7} \div 3^{-10} = 3^{-7 - (-10)} = 3^{3}$
Now multiply $3^3 \times 3^{-5}$:
$3^{3 + (-5)} = 3^{-2} = \frac{1}{3^2} = \frac{1}{9}$

(v) $2^{-3} \times (-7)^{-3}$:
$2^{-3} = \frac{1}{2^3} = \frac{1}{8}, \quad (-7)^{-3} = \frac{1}{(-7)^3} = \frac{1}{-343}$
$2^{-3} \times (-7)^{-3} = \frac{1}{8} \times \frac{1}{-343} = \frac{1}{-2744}$

3. Find the value of:

(i) $(3^0 + 4^{-1}) \times 2^2$:
$3^0 = 1, \quad 4^{-1} = \frac{1}{4}, \quad 2^2 = 4$
$(3^0 + 4^{-1}) \times 2^2 = \left(1 + \frac{1}{4}\right) \times 4 = \frac{5}{4} \times 4 = 5$

(ii) $(2^{-1} \times 4^{-1}) \div 2^{-2}$:
$2^{-1} = \frac{1}{2}, \quad 4^{-1} = \frac{1}{4}, \quad 2^{-2} = \frac{1}{2^2} = \frac{1}{4}$
$(2^{-1} \times 4^{-1}) \div 2^{-2} = \left(\frac{1}{2} \times \frac{1}{4}\right) \div \frac{1}{4} = \frac{1}{8} \div \frac{1}{4} = \frac{1}{8} \times 4 = \frac{4}{8} = \frac{1}{2}$

(iii) $\left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2}$:
Using $\left(\frac{1}{a}\right)^{-n} = a^n$:
$\left(\frac{1}{2}\right)^{-2} = 2^2 = 4, \quad \left(\frac{1}{3}\right)^{-2} = 3^2 = 9, \quad \left(\frac{1}{4}\right)^{-2} = 4^2 = 16$
$\left(\frac{1}{2}\right)^{-2} + \left(\frac{1}{3}\right)^{-2} + \left(\frac{1}{4}\right)^{-2} = 4 + 9 + 16 = 29$

Let me know if you'd like more details or help with similar problems!

Related Questions to Expand Understanding:

1. What is the difference between negative and fractional exponents?
2. How do you simplify powers when the base is a fraction?
3. What is the significance of zero exponents in algebra?
4. How do exponent rules apply to complex bases like $(-3)^4$?
5. How do you handle exponentiation in compound expressions like $(a^m b^n)^p$?

Tip:

Always simplify exponents step by step using rules like $a^m \cdot a^n = a^{m+n}$ and $(a^m)^n = a^{mn}$.