3.1 Simplify the following, without using a calculator: 3.1.1 $7(3)^0$ 3.1.2 $\\sqrt{(242 - \\sqrt{72})}$ 3.1.3 $rac{9^{n-1} \times 27^{3-2n}}{81^{2-n}}$ 3.2 Solve for $x$: $\log(x + 2) - \log x = 1$ 3.3 Given the complex number: $z = 5 - 5i$ 3.3.1 Write down the quadrant, in the complex plane, in which $z$ lies - NSC Technical Mathematics - Question 3 - 2023 - Paper 1

First, simplify $\sqrt{72}$:

$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}.$
Now compute:

$242 - 6\sqrt{2} = 2\sqrt{4 \cdot 121 - 6\cdot 1} = 2\sqrt{(121 - 6\sqrt{2})}.$

Express all numbers as powers of 3:

$\frac{(3^2)^{n-1} \times (3^3)^{3-2n}}{(3^4)^{2-n}} = \frac{3^{2(n-1)} \times 3^{9-6n}}{3^{8-4n}} = \frac{3^{2n-2 + 9 - 6n}}{3^{8-4n}}.$
Using properties of exponents, combine:

$3^{2n-2+9-6n - (8-4n)} = 3^{-4n + 7}.$
Thus, we have:

$\frac{1}{3^{4-n}} = \frac{1}{3^{(n - 7)}}.$

Using the properties of logarithms:

$\log\left(\frac{x+2}{x}\right) = 1.$
Exponential form gives:

$\frac{x+2}{x} = 10.$ Now, solving for $x$:

$x + 2 = 10x \Rightarrow 2 = 9x \Rightarrow x = \frac{2}{9}.$

The real part of $z$ is 5 (positive) and the imaginary part is -5 (negative), hence, $z$ lies in the 4th quadrant.

To express $z = 5 - 5i$ in polar form, calculate the modulus and argument:

$r = \sqrt{5^2 + (-5)^2} = \sqrt{50} = 5\sqrt{2},$
$\theta = \tan^{-1}\left(\frac{-5}{5}\right) = -\frac{\pi}{4}.$
Thus, the polar form is:

$z = 5\sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4})).$

First, substitute $z$ into the equation:

$m = 3i(5 - 5i) + 7 - ni.$
This simplifies to:

$= 15i + 15 + 7 - ni = 15 + 7 + 15i - ni = 22 + (15 - n)i.$