3.1 Simplify the following without the use of a calculator: 3.1.1 $\frac{8 x^{3} y^{2}}{16 x^{r} y^{r}}$ (leave the answer with positive exponents) 3.1.2 $\sqrt{48 + \sqrt{12}} / 27$ 3.2 If $\log 5 = m$, determine the following in terms of m: 3.2.1 $\log 25$ 3.2.2 $\log 10$ 3.3 Solve for $x$: $\log_{2} (x + 3) - 3 = -\log_{2} (x - 4)$ 3.4 Given complex numbers: $z_{1} = -1 + 3i$ and $z_{2} = \sqrt{2} \operatorname{cis} 135^{\circ}$: 3.4.1 Write down the conjugate of $z_{1}$ - NSC Technical Mathematics - Question 3 - 2022 - Paper 1

To simplify the expression:

1. Begin with simplifying $\sqrt{12}$:
   [ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} ]
2. Substitute this back in:
   [ \sqrt{48 + 2\sqrt{3}} / 27 ]
3. Now simplify $\sqrt{48}$:
   [ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} ]
4. Thus the full expression is:
   [ \sqrt{4\sqrt{3} + 2\sqrt{3}} / 27 = \sqrt{6\sqrt{3}} / 27 ]

Using the properties of logarithms:

1. Recognize that $25 = 5^{2}$:
   [ \log 25 = \log(5^{2}) = 2\log 5 ]
2. Substituting $\log 5 = m$:
   [ \log 25 = 2m ]

Using the change of base formula and properties:

1. We know $\log 10 = \log(2 \cdot 5)$:
   [ \log 10 = \log 2 + \log 5 ]
2. Thus:
   [ \log 10 = \log 2 + m ]

To solve for $x$:

1. Begin by rewriting the equation:
   [ \log_{2} (x + 3) = -\log_{2} (x - 4) + 3 ]
2. Convert the right side:
   [ \log_{2} (x + 3) = \log_{2} (\frac{1}{x - 4}) + 3 ]
3. Remove logarithm by converting to exponent:
   [ x + 3 = 2^{3} (x - 4) ]
4. Thus:
   [ x + 3 = 8(x - 4) ]
5. Expanding yields:
   [ x + 3 = 8x - 32 ]
6. Rearranging gives:
   [ 7x = 35 \Rightarrow x = 5 \text{ or } x = -4 ]

For the complex number:

1. The conjugate of $z_{1} = -1 + 3i$ is given by changing the sign of the imaginary part:
   [ \overline{z_{1}} = -1 - 3i ]

Using the polar to rectangular conversion:

1. Given $z_{2} = \sqrt{2} \operatorname{cis} 135^{\circ}$:
2. Expand it using the cosine and sine:
   [ z_{2} = \sqrt{2} \left( \cos 135^{\circ} + i \sin 135^{\circ} \right) ]
3. Thus:
   [ z_{2} = \sqrt{2}\left(-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} \right) = -1 + i ]

To evaluate:

1. Substitute the values:
   [ z_{1} = -1 + 3i \text{ and } z_{2} = -1 + i ]
2. The subtraction gives:
   [ z_{1} - z_{2} = (-1 + 3i) - (-1 + i) ]
3. Thus simplifying yields:
   [ z_{1} - z_{2} = (3i - i) = 2i ]

Rearranging gives:

1. Combine like terms:
   [ x + yi + 1 - i = 4 + 5i ]
2. This leads to:
   [ x + 1 + (y - 1)i = 4 + 5i ]
3. Equating real and imaginary parts gives two equations:
   [ x + 1 = 4 \quad\Rightarrow\quad x = 3 ]
   [ y - 1 = 5 \quad\Rightarrow\quad y = 6 ]
   Therefore:
   [ x = 3 \text{ and } y = 6 ]