Q1. a. Simplify the following expressing each answer in a power. b. Find the truth set of the equation. c. Solve the equation 2^(x-1) - 9(2^x) + 4 = 0

Let $H$ be the total height of the pylon.
Let $h$ be the height of the pylon above the theodolite's eye level.
Let $x$ be the initial horizontal distance from the theodolite to the foot of the pylon.
The height of the theodolite is $1.5\text{m}$. Therefore, $H = h + 1.5$.

From the first position, the angle of elevation is $30^\circ$:

$$\tan(30^\circ) = \frac{h}{x}$$ $$h = x \tan(30^\circ) \quad (1)$$

The surveyor moves $60\text{m}$ nearer, so the new distance is $x - 60$. The new angle of elevation is $42^\circ$:

$$\tan(42^\circ) = \frac{h}{x - 60}$$ $$h = (x - 60) \tan(42^\circ) \quad (2)$$

Equate the expressions for $h$ from equations (1) and (2):

$$x \tan(30^\circ) = (x - 60) \tan(42^\circ)$$ $$x \tan(30^\circ) = x \tan(42^\circ) - 60 \tan(42^\circ)$$

Rearrange the terms to solve for $x$:

$$60 \tan(42^\circ) = x \tan(42^\circ) - x \tan(30^\circ)$$ $$60 \tan(42^\circ) = x (\tan(42^\circ) - \tan(30^\circ))$$ $$x = \frac{60 \tan(42^\circ)}{\tan(42^\circ) - \tan(30^\circ)}$$

Using a calculator:
$\tan(30^\circ) \approx 0.57735$
$\tan(42^\circ) \approx 0.90040$

$$x = \frac{60 \times 0.90040}{0.90040 - 0.57735}$$ $$x = \frac{54.024}{0.32305}$$ $$x \approx 167.23\text{m}$$

Substitute the value of $x$ back into equation (1):

$$h = x \tan(30^\circ)$$ $$h = 167.23 \times 0.57735$$ $$h \approx 96.52\text{m}$$

The total height of the pylon $H$ is the height $h$ plus the height of the theodolite ($1.5\text{m}$):

$$H = h + 1.5\text{m}$$ $$H = 96.52\text{m} + 1.5\text{m}$$ $$H = 98.02\text{m}$$

The height of the pylon is $\boxed{\text{98.02 m}}$.