3.1 Vereenvoudig die volgende SONDERS om 'n sakrekenaar te gebruik: 3.1.1 $log 3 + log 27$ $log 81 - log 9$ 3.1.2 \[ \frac{2^{\sqrt{32}} + 2^{\sqrt{2}}}{2^{\sqrt{50}}} \] 3.2 Los op vir $x$: $log_{32} + log 4 - log_{16} = log_{125}$ 3.3 Twee wisselstroombane, wat in serie verbind is, het impedansies $Z_1 = 4 + 5i$ en $Z_2 = 4 - 4i$ - NSC Technical Mathematics - Question 3 - 2020 - Paper 1

We simplify the expression: [ \frac{2^{\sqrt{32}} + 2^{\sqrt{2}}}{2^{\sqrt{50}}} = \frac{2^{2 \cdot 4} + 2^{\sqrt{2}}}{2^{\sqrt{50}}} = \frac{2^{4} + 2^{\sqrt{2}}}{2^{\sqrt{50}}} = 1]

Using logarithm properties: $log_{32}(2^5) + log(2^2) - log(2^4) = log(125)$ This can further be simplified using change of base: $\frac{log(32) + log(4) - log(16)}{log(2)} = log(125)$ After simplification, we find that $x = 2$.

To find the total impedance: $Z_T = Z_1 + Z_2 = (4 + 5i) + (4 - 4i) = 8 + i$

To express the total impedance in polar form, we calculate the modulus: $r = |Z_T| = \sqrt{8^2 + 1^2} = \sqrt{64 + 1} = \sqrt{65}$

Next, we find the angle: $\theta = tan^{-1}\left(\frac{1}{8}\right)$

Finally, we express it as: $Z_T = r(cos \theta + i sin \theta)$

First, simplify the expression: $k = 6 + 4(i - 9) + 2mi = 6 + 4i - 36 + 2mi = -30 + (4 + 2m)i$ Setting the real part and imaginary part: Real part: $-30$ Imaginary part: $4 + 2m$

This leads us to the final values for $k$.