3.1 Vereenvoudig: $(-2 \\sqrt{a})^8$ $=(2^8)(a)^{\frac{4}{3}}$ $=256a^8$ 3.2 Los op vir $x$: $ ext{log}_3(3x-2) + \text{log}_{0.5} = 3$ $ ext{log}_3(3x-2) = 3 - \text{log}_{0.5}$ $ ext{log}_3(3x-2) = 2^3 \\\text{(since } \text{log}_{0.5} = -1)$ $3x - 2 = 2^2$ $3x = 4 + 2$ $3x = 6$ $x = 2$ 3.3 Indien log $2 = a$ en log $3 = b$, bepaal die waarde van log $\ ext{log} \rac{1}{\sqrt{6}}$ in terme van $a$ en $b$ - NSC Technical Mathematics - Question 3 - 2019 - Paper 1

Combine logarithmic terms.

1. Rearrange:
   $\text{log}_3(3x-2) = 3 - \text{log}_{0.5}$
   2. Substitute:
   $\text{log}_3(3x-2) = 3 + 1$
   3. Solve for $x$:
   $3x - 2 = 2^6$ $3x = 64 + 2$ $3x = 66$ $x = 22$.

1. Express in terms of base:
   $\log \sqrt{6} = \frac{1}{2}\log (6)$
   2. Substitute the logs:
   $= \frac{1}{2} (\log(2) + \log(3))$
   $= \frac{1}{2} (a + b)$.

Use the formula for voltage given in polar form:
$V = r(cos\theta + i sin\theta)$
Substitute $r = 2$ and $heta = 240^{\circ}$:
$V = 2(cos(240^{\circ}) + i sin(240^{\circ}))$
Evaluate:
$= 2(\frac{-1}{2} + i \frac{\sqrt{3}}{2})$
$= -1 + \sqrt{3}i$.