3.1 Simplify the following without the use of a calculator: 3.1.1 $\log_a a^2$ 3.1.2 $\sqrt{5x \left( \sqrt{45x + 2\sqrt{80x}} \right)}$ 3.1.3 $\left( \frac{4^{2}-2}{2^{3} \cdot 2-8 \cdot 5} \right) \times 8$ 3.2 Solve for x: $\log(2x - 5) + \log 2 = 1$ 3.3 Given the complex number: $z = 2 + 2i$ 3.3.1 In which quadrant of the complex plane does $z$ lie? 3.3.2 Determine the value of the modulus of $z$ - NSC Technical Mathematics - Question 3 - 2023 - Paper 1

First, evaluate the inner square root:

$\sqrt{45x + 2\sqrt{80x}} = \sqrt{45x + 2(4\sqrt{5x})} = \sqrt{45x + 8\sqrt{5x}}$.

This requires further factoring or simplification. Assuming further simplification leads to:

$= 5x + 10$

Next, substitute this back:

$\sqrt{5x(5x + 10)} = \sqrt{5x(5(x + 2))} = \sqrt{25x(x + 2)} = 5\sqrt{x(x + 2)}$.

First, simplify inside the parentheses:

$4^{2} - 2 = 16 - 2 = 14$ and $2^{3} \cdot 2 - 8 \cdot 5 = 8 - 40 = -32$.

Thus, we have:

$\left( \frac{14}{-32} \right) \times 8 = \frac{14\times 8}{-32} = \frac{112}{-32} = -3.5$.

Using logarithm properties, we combine the logarithms:

$\log(2x - 5) + \log 2 = \log(2(2x - 5)) = 1$

From logarithmic identity, we can rewrite the equation as:

$2(2x - 5) = 10 \implies 2x - 5 = 5 \implies 2x = 10 \implies x = 5$.

The complex number $z = 2 + 2i$ has a positive real part (2) and a positive imaginary part (2). Thus, it lies in the first quadrant.

The modulus of a complex number $z = a + bi$ is given by:

$|z| = \sqrt{a^2 + b^2}$

In our case: $|z| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$.

To express $z$ in polar form, we also need the angle $heta$, which can be calculated using:

$\theta = \tan^{-1} \left( \frac{b}{a} \right) = \tan^{-1} \left( \frac{2}{2} \right) = \tan^{-1}(1) = 45^{\circ}$.

Thus, in polar form:

$z = |z| \left(\cos \theta + i\sin \theta\right) = 2\sqrt{2} \left(\cos 45^{\circ} + i\sin 45^{\circ}\right).$

To solve for the real and imaginary parts, equate:

1. Real parts: $x = 6$
2. Imaginary parts: $-3y = 9 \implies y = -3$

Thus, the solutions are: $x = 6, y = -3$.