3.1 Simplify the following WITHOUT using a calculator: 3.1.1 - NSC Technical Mathematics - Question 3 - 2021 - Paper 1

Using the properties of logarithms:

( \log_{2}16 = \log_{2}(2^4) = 4 )
( \log_{4}4 = \log_{4}(4^1) = 1 )

Thus: $\log_{2}16 + \log_{4}4 = 4 + 1 = 5$

First, simplify each square root:

$\sqrt{50 \cdot 10^{6}} = \sqrt{50} \cdot \sqrt{10^{6}} = \sqrt{50} \cdot 100 = 10\sqrt{50}$ The value of ( \sqrt{50} = 5\sqrt{2} ) So: $= 10 \cdot 5 \cdot 100 = 500\sqrt{2}$

Now for the second part: $\sqrt{18 \cdot 4} = \sqrt{72} = 6\sqrt{2}$

Combining both results: $500\sqrt{2} \cdot 6\sqrt{2} = 3000.$

Use the property of logarithms: $\log_{3}(x + 2) - \log_{3}x = 2$

Combining: $\log_{3}\left( \frac{x + 2}{x} \right) = 2$

Exponentiating gives: $\frac{x + 2}{x} = 3^{2} = 9$

Thus: $x + 2 = 9x$ $2 = 8x \Rightarrow x = \frac{1}{4}$

Given the modulus: $|z| = \sqrt{p^2 + 4^2} = \frac{\sqrt{5}}{2}$

Squaring both sides: $p^2 + 16 = \frac{5}{4}$

Thus: $p^2 = \frac{5}{4} - 16 = \frac{5 - 64}{4} = \frac{-59}{4}$

Since this is not possible, review the argument to derive a suitable $p$ within the specified range.

In polar form:
Given $|z| = \frac{\sqrt{5}}{2}$ and argument $\theta \in (90^{\circ}, 180^{\circ})$,

The polar representation is: $z = r \text{cis} \theta = \frac{\sqrt{5}}{2} \text{cis} \theta$ where $\theta$ can be calculated to be in radians.

First, distribute on the right side: $-3i(4 + 5) = -12i - 15i = -27i$

Now we have: $2m - ni - 6i = -27i$

This leads to: $2m = -27i + (n + 6)i.$

Equate imaginary parts: $2m + 6 = -27$

Solving gives: $m = -\frac{33}{2}, n = 9$