3.1 Vereenvoudig die volgende sonder om 'n sakrekenaar te gebruik: 3.1.1 $\log_a a^{1} $ 3.1.2 $ \sqrt{5x \left( \sqrt{45x + 2 + \sqrt{80x}} \right)} $ 3.1.3 $ \left( \frac{4^{x-2}}{2^{2x-3} - 8^{3}} \right) \times 8 $ 3.2 Los op vir $x$: $\log(2x - 5) + \log 2 = 1 $ 3.3 Gegee die komplekse getal: $z = 2 + 2i $ 3.3.1 In watter kwadrant van die komplekse vlak lê $z$? 3.3.2 Bepaal die waarde van die modulus van $z$ - NSC Technical Mathematics - Question 3 - 2023 - Paper 1

Start by simplifying the expression inside the square root:

First, compute $\sqrt{80x}$:

$\sqrt{80x} = \sqrt{16 \cdot 5x} = 4\sqrt{5x}$

Now substitute this back into the expression:

$\sqrt{5x (\sqrt{45x + 2 + 4\sqrt{5x}})}$

This can be simplified as:

$= \sqrt{5x (\sqrt{5(9x + 0.4 + 4\sqrt{x})})}$

Further simplification leads us to:

$= \sqrt{45x + 2 + 4\sqrt{5x}}$

First, rewrite $8$ as $2^3$:

$= \left( \frac{4^{x-2}}{2^{2x-3} - 2^{3}} \right) \times 2^3$

Now apply the power rule:

$4^{x-2} = (2^2)^{x-2} = 2^{2(x-2)} = 2^{2x-4}$

Substituting back,

$= \frac{2^{2x-4}}{2^{2x-3} - 2^{3}} \times 2^{3}$

This leads us to simplify the expression.

Combine the logarithmic terms:

$\log(2x - 5) + \log(2) = \log(2(2x - 5))\text{ (using product property)}$

Setting it equal to 1:

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

Converting to exponential form gives:

$2(2x - 5) = 10$

Solving for $x$:

$2x - 5 = 5\implies 2x = 10\implies x = 5$

The complex number $z = 2 + 2i$ lies in the first quadrant because both its real part (2) and its imaginary part (2) are positive.

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

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

For $z = 2 + 2i$:

$|z| = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

To express $z$ in polar form, we need the modulus and the argument (angle). The argument is given by:

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

For $z = 2 + 2i$:

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

Thus, in polar form:

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

To solve for $x$ and $y$, separate the equation into real and imaginary parts:

$x - 3y = 6 + 9i$

Thus,

Real part: $x - 3y = 6$

Imaginary part: $0 = 9$ (not applicable if treated correctly)

From the first equation:

$x = 6 + 3y$

Therefore, without a second equation available in this context, we can express $x$ in terms of $y$.