Simplify the following WITHOUT using a calculator: 3.1.1 $(8^{1-a^{-3}})^{\frac{3}{4}}$ 3.1.2 $ ext{log}_2 16 + ext{log}_4 4$ 3.1.3 $\sqrt{50x^{10}} \cdot \sqrt{18x^{-4}}$ 3.2 Solve for $x$: $ ext{log}_3 (x+2) = 2 + \text{log}_3 x$ 3.3 Given the complex number $z = p + 4i$ with the modulus $\frac{2}{\sqrt{5}}$ and the argument $\theta \in (90^{\circ}, 180^{\circ})$ - NSC Technical Mathematics - Question 3 - 2021 - Paper 1

Using the properties of logarithms, we simplify:

1. Start with $ext{log}_2 16$: $\text{log}_2 16 = \text{log}_2 (2^4) = 4$

2. Calculate $ext{log}_4 4$: $\text{log}_4 4 = \text{log}_4 (4^1) = 1$

3. Combine the results: $4 + 1 = 5$

To simplify this product of square roots, apply the property $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$:

$\sqrt{50x^{10}} \cdot \sqrt{18x^{-4}} = \sqrt{50 \cdot 18 \cdot x^{10} \cdot x^{-4}}$

First, simplify the product inside the square root: $50 \cdot 18 = 900 \quad \text{and} \quad x^{10} \cdot x^{-4} = x^{10-4} = x^{6}$

Thus: $= \sqrt{900x^{6}} = \sqrt{900} \cdot \sqrt{x^{6}} = 30x^{3}$

To solve for $x$, start by isolating the logarithmic expressions:

1. Rewrite $2$ as $ext{log}_3 9$ (since $3^2 = 9$): $\text{log}_3 (x+2) = \text{log}_3 9 + \text{log}_3 x$

2. Combine the logarithms using the property of logarithms: $\text{log}_3 (x + 2) = \text{log}_3 (9x)$

3. Set the arguments equal to each other since the logarithm is one-to-one: $x + 2 = 9x$

4. Solve for $x$: $2 = 9x - x\implies 2 = 8x \implies x = \frac{1}{4}$

Given that the modulus is $\frac{2}{\sqrt{5}}$, we use the formula for modulus:

$|z| = \sqrt{p^2 + 4^2}$

Set this equal to the given modulus: $\sqrt{p^2 + 16} = \frac{2}{\sqrt{5}}$

Square both sides: $p^2 + 16 = \left(\frac{2}{\sqrt{5}}\right)^2$ $p^2 + 16 = \frac{4}{5}$

Now, isolate $p^2$: $p^2 = \frac{4}{5} - 16 = \frac{4 - 80}{5} = \frac{-76}{5}$

Since $p^2$ cannot be negative, there's a misinterpretation of the parameters in the problem. We need additional context for $p$.

The polar form of a complex number is expressed as $z = r \text{cis} \theta$, where $r$ is the modulus and $heta$ is the argument.

1. From our previous calculation, assume we've defined $r$ correctly.
2. The argument is given as $\theta \in (90^{\circ}, 180^{\circ})$.
3. Thus, with placeholder values assuming we find correct $p$ and $\theta$: $z = r \text{cis} \theta$

Final form is expressed as specified based on modulus and angle.

To solve, start with the equation: $2m - ni - 6i = -3i(4 + 5)$

1. Compute the right side: $-3i(4 + 5) = -3i \cdot 9 = -27i$

2. Rewrite the equation: $2m - ni - 6i = -27i$

3. Combine like terms: $2m - (n + 6)i = -27i$

Equating real and imaginary parts gives:

1. Real part: $2m = 0\implies m = 0$,
2. Imaginary part: $-(n + 6) = -27\implies n + 6 = 27\implies n = 21$.