3.1 Simplify the following WITHOUT using a calculator: 3.1.1 $\frac{3^3 \times 3^{-2}}{9^1}$ 3.1.2 $\frac{\sqrt{5+4}-\sqrt{-45}}$ 3.1.3 $log_3 8 + log_{10} 25$ 3.2 Solve for $x$: $log_4 (x-6) = log_25$ 3.3 In the RLC circuit, the impedance of the two impedances connected in series are: $Z_1 = 4/2 \text{cis} 225^\circ$ and $Z_2 = 3 - 4i$ 3.3.1 Express $Z_1$ in rectangular form - NSC Technical Mathematics - Question 3 - 2022 - Paper 1

First, simplify the first term:

$\sqrt{5+4} = \sqrt{9} = 3$

For the second term, since we are dealing with a negative under the square root, we can convert it to:

$\sqrt{-45} = \sqrt{45}i = 3\sqrt{5}i$

Combining both, we have:

$3 - 3\sqrt{5}i$

Using the change of base formula, we rewrite:

$log_3 8 = \frac{log_2 8}{log_2 3} = \frac{3}{log_2 3}$

For the second term:

$log_{10} 25 = log_{10} (5^2) = 2 \cdot log_{10} 5$

Thus, our final expression combining both logs becomes:

$\frac{3}{log_2 3} + 2 \cdot log_{10} 5$

Firstly, convert the logarithm of base 4 to base 2:

$log_4 (x-6) = \frac{log_2 (x-6)}{log_2 4} = \frac{log_2 (x-6)}{2}$

Setting this equal to $log_2 25$ gives:

$\frac{log_2 (x-6)}{2} = log_2 25$

Multiplying everything by 2:

$log_2 (x-6) = 2 \cdot log_2 25$

Now, simplify:

$x-6 = 25^2 = 625 \Rightarrow x = 625 + 6 = 631$

To express $Z_1 = 4/2 \text{cis} 225^\circ$, we need to convert to rectangular form:

$Z_1 = \frac{4}{2} (cos(225^\circ) + i sin(225^\circ)) = 2(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i) = -\sqrt{2} - \sqrt{2}i$

Now, adding $Z_1$ and $Z_2$:

$Z_2 = 3 - 4i$

Thus:

$Z_1 + Z_2 = (-\sqrt{2} - \sqrt{2}i) + (3 - 4i)$

Combine real and imaginary parts:

$= (3 - \sqrt{2}) + (-4 - \sqrt{2})i$

The total impedance becomes:

$Z = (3 - \sqrt{2}) + (-4 - \sqrt{2})i$

First, simplify $4i^{-2}$:

$4i^{-2} = 4\cdot(-1) = -4$

Now, substituting:

$-p + qi = -4 -(2 + 3i)$

Combine terms:

$-p + qi = -6 - 3i$

By matching real and imaginary parts:

$-p = -6 \Rightarrow p = 6$ $q = -3$