3.1 Simplify (showing ALL calculations) the following WITHOUT using a calculator: 3.1.1 $\\sqrt{16 a^6}$ 3.1.2 $\\\sqrt{\\log_{32} + \log 100 + 9}$ 3.1.3 $(\\4\\sqrt{5} + \sqrt{2})(\\\sqrt{2 - 4\\sqrt{5}})$ 3.2 Solve for $x$: $\\log_{x} x = 3 - \log_{x} (x + 6)$ 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 $z = r cis \theta$ (where $ heta$ is in degrees) 3.4 Solve for $a$ and $b$ if $a + b + (a + bi) - bi = 5 - 3i$ - NSC Technical Mathematics - Question 3 - 2021 - Paper 1

First, we use the property of logarithms:

$\\log_{32} = \frac{\\log_{10} 32}{\\log_{10} 10} = \frac{5}{2}.$

Next, we find:

$\\log 100 = 2.$

Now combine:

$\\\log_{32} + \log 100 + 9 = \frac{5}{2} + 2 + 9 = \frac{5}{2} + \frac{4}{2} + \frac{18}{2} = \frac{27}{2}.$

Taking the square root, we find:

$\\\sqrt{\\\frac{27}{2}} = \frac{3\\\sqrt{3}}{2}.$

Thus, the final result is:

$\boxed{\\\frac{3\sqrt{3}}{2}}$

Start with the expression:

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

Please note that:

$\\2 - 4\\sqrt{5}$ is negative, so we can rewrite it using:

$\\\sqrt{x^2} = |x|$

due to the quadratic nature of the expression. Thus:

The product yields a complex expression that cannot be simplified further clearly without specific values for roots derived from actual numerical values. It simplifies logarithmically depending on conditions.

Thus:

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

Using properties of logarithms:

We rewrite the equation as:

$\\log_{x} x + \log_{x} (x + 6) = 3.$

Then we know:

$\\log_{x} (x(x + 6)) = 3$

Which converts to:

$x(x + 6) = x^3 = 27.$

So, the resultant cubic equation:

$x^2 + 6x - 27 = 0$

Applying the quadratic formula:

= \frac{-6 \pm 12}{2}$$ Thus: $$x = 3\text{ or } x = -9.$$ The valid solution is: $$\boxed{3}$$

Given the relation:

$z = 2w - 7i$

and substituting for $w$:

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

Calculating gives:

$= (1 + 6i) - 7i = 1 - i.$

Therefore, we have:

$z = \boxed{1 - i}$

To convert to polar form, first calculate the modulus:

$r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (-1)^2} = \sqrt{2}.$

Next, compute the argument:

$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-1}{1}\right) = 315^\circ.$

Thus, in polar form:

$z = \sqrt{2} \text{cis} 315^\circ.$

Hence, our final result is:

$\boxed{z = \sqrt{2} \text{cis} 315^\circ}$

Starting with:

$a + b + (a + bi) - bi = 5 - 3i$

Combine like terms:

$2a + b = 5$ $\text{and for } i: 0 = -3 + b = -3$

This yields $b = -3.$

Now substitute back to find $a$:

2a = 8 \\ a = 4.$$ Thus the values are: $$\boxed{a = 4, b = -3}$$