3.1 Simplify (showing ALL calculations) the following WITHOUT using a calculator: 3.1.1 $\\sqrt{16a^{6}}$ 3.1.2 $\rac{\\log, 32 + \log 100 + 9}{\\log, 32 - \log 100 + 9}$ 3.1.3 $(4\sqrt{5} + \sqrt{2}) (\sqrt{2 - 4\sqrt{5}})$ 3.2 Solve for $x$: $\log_{10}x = 3 - \log_{10}(x + 6)$ 3.3 Given the complex number $z = 2w - 7i$ where $w = \frac{1}{2} + 3i$ 3.3.1 Determine $z$ in the form $a + bi$ 3.3.2 Express $z$ in the polar form $z = r \text{cis } \theta$ (where $\theta$ is in degrees) 3.4 Solve for $a$ and $b$ if $a + b + ai - bi = 5 - 3i$ - NSC Technical Mathematics - Question 3 - 2021 - Paper 1

We start by using the logarithm properties:

Recall that: $\log a + \log b = \log(ab)$ and $\log a - \log b = \log(\frac{a}{b})$.

First compute the numerator and denominator separately:

Numerator: $\log(32) + \log(100) + 9 = \log(32 \times 100) + 9$ $= \log(3200) + 9$

Denominator: $\log(32) - \log(100) + 9 = \log(\frac{32}{100}) + 9$ $= \log(0.32) + 9$

Thus, we have: $\frac{\log(3200) + 9}{\log(0.32) + 9}$

This expression involves distributing the terms.

First, compute $\sqrt{2 - 4\sqrt{5}}$.

Let's simplify the expression inside the square root:

Calculate: $2 - 4\sqrt{5}$

Assuming further simplification can be done, we multiply:

$(4\sqrt{5} + \sqrt{2})(\sqrt{2 - 4\sqrt{5}})$

This will not yield simple results without further information, so we would leave it as is unless specific values are provided.

To solve for $x$, we first manipulate the equation.

Rearranging gives: $\log_{10}x + \log_{10}(x + 6) = 3$

Using the property of logarithms: $\log_{10}(x(x + 6)) = 3$

Exponentiating both sides: $x(x + 6) = 10^{3}$ $x(x + 6) = 1000$

This leads to the quadratic equation: $x^{2} + 6x - 1000 = 0$

Apply the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}$ Here: $a = 1$, $b = 6$, and $c = -1000$.

Calculating the discriminant: $b^2 - 4ac = 36 + 4000 = 4036$

Thus: $x = \frac{{-6 \pm \sqrt{4036}}}{2}$

Finally, calculate the roots to find $x$.

Given the expression $z = 2w - 7i$ and $w = \frac{1}{2} + 3i$, we substitute:

$z = 2\left(\frac{1}{2} + 3i\right) - 7i$

Calculating: $= 1 + 6i - 7i$ $= 1 - i$

Thus, in the form $a + bi$, we have: $z = 1 - i$

For the complex number $z = 1 - i$, we first find the modulus $r$:

$r = \sqrt{a^2 + b^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}$

Next, we determine the angle $\theta$ using the tangent function:

$\tan \theta = \frac{b}{a} = \frac{-1}{1}$

This gives: $\theta = \text{tan}^{-1}(-1)$

In degrees, this is $315^{\circ}$ (fourth quadrant).

Thus, the polar form is: $z = \sqrt{2} \text{cis } 315^{\circ}$

To solve for $a$ and $b$, we first rewrite the equation:

$a + b + ai - bi = 5 - 3i$

Grouping real and imaginary parts gives: $a + b = 5$ $a - b = -3$

Now, we can solve this system of equations. Adding the two equations: $2a = 2 \implies a = 1$

Substituting $a$ back into one of the equations, say $a + b = 5$: $1 + b = 5 \implies b = 4$

Therefore, the solution is: $a = 1, b = 4$