Vereenvoudig die volgende sonder om 'n sakrekenaar te gebruik: 3.1.1 $rac{8 x^3 y^2}{16 x^6 y^4}$ (laat die antwoord met positiewe eksponente) 3.1.2 $rac{ ext{√}(48 + ext{√}12)}{27}$ Indien $log_5 m$, bepaal die volgende in terme van $m$: 3.2.1 $log 25$ 3.2.2 $log 2$ Los op vir $x$: $log_2 (x + 3) - 3 = -log_2 (x - 4)$ Gegge die komplekse getalle: $z_1 = -1 + 3i$ en $z_2 = ext{√}2 ext{cis} 135^{ ext{o}}$ 3.4.1 Skryf die toegemaakte van $z_1$ neer - NSC Technical Mathematics - Question 3 - 2022 - Paper 1

First, simplify the expression inside the square root:

$48 = 16 imes 3 ext{ and } ext{√}12 = 2 ext{√}3$

So,

$ext{√}(48 + ext{√}12) = ext{√}(16 imes 3 + 2 ext{√}3)$

You can factor out terms further to simplify, but for clarity, we shall keep this result as:

rac{ ext{√}(48 + 2 ext{√}3)}{27}

Since $log_5 m$, we can rewrite:

$log 25 = log(5^2) = 2 log 5 = 2m$

Using the change of base property:

$log 2 = log(10) - log(5) = 1 - m$

Rearranging gives:

$log_2 (x + 3) + log_2 (x - 4) = 3$

This leads to:

$log_2((x + 3)(x - 4)) = 3$

Thus:

$(x + 3)(x - 4) = 2^3 = 8$

Expanding the left side, we have:

ightarrow x^2 - x - 20 = 0$$ Using the quadratic formula: $$x = rac{-(-1) ext{±} ext{√}((-1)^2 - 4(1)(-20))}{2(1)}$$ This yields $x = 5$ or $x = -4$.

The conjugate of $z_1 = -1 + 3i$ is:

$z_1^* = -1 - 3i$

We expand:

$z_2 = ext{√}2 ext{cis} 135^{ ext{o}} = ext{√}2 ( ext{cos} 135^{ ext{o}} + i ext{sin} 135^{ ext{o}})$

Calculating:

= ext{√}2(-rac{1}{ ext{√}2} + i imes rac{1}{ ext{√}2}) = -1 + i

Thus, using the values found:

$z_1 - z_2 = (-1 + 3i) - (-1 + i) = 2i$

Rearranging gives:

$x + yi = 4 + 5i + 1 - i$

Thus:

$x + yi = 5 + 4i$

From here, we identify:

$x = 5 ext{ and } y = 4$