3.1 Simplify the following without the use of a calculator: 3.1.1 $\\sqrt{8x^{27}}$ 3.1.2 $9^{1} \cdot 4^{1} \cdot 6^{-2n}$ 3.1.3 $\sqrt{2 - \sqrt{-4}} - \sqrt{4k}$ 3.2 Given: $rac{\log 72 - \log 2}{\log 6}$ 3.2.1 Write the following as a single logarithm: $\\log 72 - \log 2$ 3.2.2 Hence, simplify without using a calculator: $rac{\log 72 - \log 2}{\log 6}$ 3.3 Solve for $x$: $5^{x^{2} - 5^{2}} = 600$ 3.4 Given the complex numbers: $r_{1} = 2 + 3i$ and $r_{2} = i$ 3.4.1 Write down the conjugate of $r_{2}$ - NSC Technical Mathematics - Question 3 - 2024 - Paper 1

First, we convert the bases to powers of primes:

$9^{1} = 3^{2}, \quad 4^{1} = 2^{2}, \quad 6^{-2n} = (2 \cdot 3)^{-2n} = 2^{-2n} \cdot 3^{-2n}$

Now we can combine:

$\Rightarrow 3^{2} \cdot 2^{2} \cdot 2^{-2n} \cdot 3^{-2n} = 2^{2 - 2n} \cdot 3^{2 - 2n}$

Thus, the simplified expression is $2^{2 - 2n} \cdot 3^{2 - 2n}$.

First, simplify $\sqrt{-4}$: $\sqrt{-4} = \sqrt{4} i = 2i$

Now substitute back: $\sqrt{2 - 2i} - \sqrt{4k}$ We can simplify further by keeping it as is for clarity unless specific values for k are given.

Using the property of logarithms that states $\log a - \log b = \log \left( \frac{a}{b} \right)$: $\log 72 - \log 2 = \log \left( \frac{72}{2} \right) = \log 36.$

We have already established that $\log 72 - \log 2 = \log 36$. Thus: $\frac{\log 36}{\log 6} = \log_{6} 36.$ Using the change of base formula $\log_{b}(a) = \frac{\log(a)}{\log(b)}$ gives us: $\log_{6}(6^2) = 2.$ Therefore, $\frac{\log 72 - \log 2}{\log 6} = 2.$

Start by rewriting the equation: $x^{2} - 25 = \log_{5}(600)$ To find solutions: $x^{2} = \log_{5}(600) + 25$ $x = \pm \sqrt{\log_{5}(600) + 25}.$

The conjugate of a complex number $a + bi$ is $a - bi$. Hence, the conjugate of $r_{2} = i$ is $-i$.

We have: $\frac{r_{1}}{r_{2}} = \frac{2 + 3i}{i}.$ Multiply the numerator and denominator by the conjugate of the denominator: $= \frac{(2 + 3i)(-i)}{i(-i)} = \frac{-2i - 3i^2}{-1} = -2i + 3 = 3 - 2i.$

Comparing real and imaginary parts gives: $a = -14, \quad b = -1.$