3.1 Vereenvoudig die volgende, sonder om 'n sakrekenaar te gebruik: 3.1.1 7(3x)º 3.1.2 √(√242 – √72) 3.1.3 9¹⁻¹ × 27³⁻²n⁻² 8¹²⁻² 3.2 Los op vir x: log(x + 2) – log x = 1 3.3 Gee die komplekse getal: z = 5 – 5i 3.3.1 Skryf die kwadrant in die kompleksvlak neer waarin z lê - NSC Technical Mathematics - Question 3 - 2023 - Paper 1

First, calculate the components:

$$√242 = √(4 imes 61) = 2√61$$

and

$$√72 = √(36 imes 2) = 6√2$$

Now subtract these values:

√(2√61 - 6√2)\

Using the exponent properties:

9^{-1} = rac{1}{9} = 3^{-2}, \ 27^{-2} = 3^{-6} \ \Rightarrow \ (3^{-2}) imes (3^{-6}) = 3^{-2 - 6} = 3^{-8}

Thus, we have:

\frac{1}{3^8} = \frac{1}{6561}\

Using the properties of logarithms, we can express this as:

$$log\left(\frac{x+2}{x}\right) = 1$$

This leads to:

$$\frac{x+2}{x} = 10\ \Rightarrow \ x + 2 = 10x\ \Rightarrow \ 9x = 2\ \Rightarrow \ x = \frac{2}{9}$$

To determine the quadrant, we observe the position of the real and imaginary parts:

• Real part: 5 > 0
• Imaginary part: -5 < 0

Thus, the complex number lies in the fourth quadrant.

The polar form of a complex number is given by: $z = r(\cos(\theta) + i\sin(\theta))$ where:

• Modulus: $r = \sqrt{(5)^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2}$
• Argument: $\theta = \tan^{-1}\left(\frac{-5}{5}\right) = -\frac{\pi}{4}$

Therefore, the polar form is: $z = 5\sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right)$

First, simplify the expression:

$$3i(2i - 5) = 3i(2i) - 3i(5) = -6 + 15i$$

Substituting into the original equation:

$$m = -6 + 15i + 7 - ni\ = 1 + 15i - ni.$$

Hence, we can separate the real and imaginary parts to find m and n.