The problem asks to solve several exponential equations. We will solve equations (a), (b), (c), (d), (e), (g), and (h) from question 1.

1. Problem Description

The problem asks to solve several exponential equations. We will solve equations (a), (b), (c), (d), (e), (g), and (h) from question

1.

2. Solution Steps

(a) $4^x = 0.5$

$4^x = \frac{1}{2}$

$(2^2)^x = 2^{-1}$

$2^{2x} = 2^{-1}$

$2x = -1$

$x = -\frac{1}{2}$

(b) $9^{x-1} = \frac{1}{729}$

$9^{x-1} = \frac{1}{9^3}$

$9^{x-1} = 9^{-3}$

$x-1 = -3$

$x = -2$

(c) $\frac{1}{5} \times 32^{x-\frac{1}{2}} = 0.1$

$\frac{1}{5} \times 32^{x-\frac{1}{2}} = \frac{1}{10}$

$32^{x-\frac{1}{2}} = \frac{5}{10} = \frac{1}{2}$

$(2^5)^{x-\frac{1}{2}} = 2^{-1}$

$2^{5(x-\frac{1}{2})} = 2^{-1}$

$5(x-\frac{1}{2}) = -1$

$5x - \frac{5}{2} = -1$

$5x = \frac{5}{2} - 1 = \frac{3}{2}$

$x = \frac{3}{10}$

(d) $125^{x-1} = 25^{2x-3}$

$(5^3)^{x-1} = (5^2)^{2x-3}$

$5^{3(x-1)} = 5^{2(2x-3)}$

$3(x-1) = 2(2x-3)$

$3x - 3 = 4x - 6$

$x = 3$

(e) $9^{x-1} = 27^{x+1}$

$(3^2)^{x-1} = (3^3)^{x+1}$

$3^{2(x-1)} = 3^{3(x+1)}$

$2(x-1) = 3(x+1)$

$2x - 2 = 3x + 3$

$x = -5$

(g) $9^{2x-1} = \frac{81^{x-1}}{3^x}$

$(3^2)^{2x-1} = \frac{(3^4)^{x-1}}{3^x}$

$3^{2(2x-1)} = \frac{3^{4(x-1)}}{3^x}$

$3^{4x-2} = 3^{4x-4-x}$

$3^{4x-2} = 3^{3x-4}$

$4x - 2 = 3x - 4$

$x = -2$

(h) $8^{\frac{x}{2}} = 2^{\frac{3}{8}} \times 4^{\frac{3}{4}}$

$(2^3)^{\frac{x}{2}} = 2^{\frac{3}{8}} \times (2^2)^{\frac{3}{4}}$

$2^{\frac{3x}{2}} = 2^{\frac{3}{8}} \times 2^{\frac{6}{4}}$

$2^{\frac{3x}{2}} = 2^{\frac{3}{8} + \frac{3}{2}}$

$\frac{3x}{2} = \frac{3}{8} + \frac{12}{8} = \frac{15}{8}$

$3x = \frac{30}{8} = \frac{15}{4}$

$x = \frac{5}{4}$

3. Final Answer

(a) $x = -\frac{1}{2}$

(b) $x = -2$

(c) $x = \frac{3}{10}$

(d) $x = 3$

(e) $x = -5$

(g) $x = -2$

(h) $x = \frac{5}{4}$