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3.1 Vereenvoudig: $$ \left( -2 \sqrt{a} \right)^8 $$ 3.2 Los op vir $x$: $$ \log_3(3x-2) + \log_0,5 = 3 $$ 3.3 Indien \( \log_2 a = b \) en \( \log_3 5 = c \), bepaal die waarde van \( \log \sqrt{0,6} \) in terme van a en b - NSC Technical Mathematics - Question 3 - 2019 - Paper 1

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3.1-Vereenvoudig:--$$-\left(--2-\sqrt{a}-\right)^8-$$--3.2-Los-op-vir-$x$:---$$-\log_3(3x-2)-+-\log_0,5-=-3-$$--3.3-Indien-\(-\log_2-a-=-b-\)-en-\(-\log_3-5-=-c-\),-bepaal-die-waarde-van-\(-\log-\sqrt{0,6}-\)-in-terme-van-a-en-b-NSC Technical Mathematics-Question 3-2019-Paper 1.png

3.1 Vereenvoudig: $$ \left( -2 \sqrt{a} \right)^8 $$ 3.2 Los op vir $x$: $$ \log_3(3x-2) + \log_0,5 = 3 $$ 3.3 Indien \( \log_2 a = b \) en \( \log_3 5 = c \), ... show full transcript

Worked Solution & Example Answer:3.1 Vereenvoudig: $$ \left( -2 \sqrt{a} \right)^8 $$ 3.2 Los op vir $x$: $$ \log_3(3x-2) + \log_0,5 = 3 $$ 3.3 Indien \( \log_2 a = b \) en \( \log_3 5 = c \), bepaal die waarde van \( \log \sqrt{0,6} \) in terme van a en b - NSC Technical Mathematics - Question 3 - 2019 - Paper 1

Step 1

Vereenvoudig: \( \left( -2 \sqrt{a} \right)^8 \)

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Answer

Om ( \left( -2 \sqrt{a} \right)^8 ) te vereenvoudig, gebruik ons die eksponentiële eienskappe:

(2a)8=(2)8(a)8=256a4.\left( -2 \sqrt{a} \right)^8 = (-2)^8 \cdot (\sqrt{a})^8 = 256 a^4.

Die vereenvoudigde vorm is dus ( 256 a^4 ).

Step 2

Los op vir $x$: \( \log_3(3x-2) + \log_{0.5} = 3 \)

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Answer

Begin met die logaritmiese vergelyking:

log3(3x2)+log0.5=3 log0.5=1,solog3(3x2)1=3log3(3x2)=43x2=343x2=813x=83x=833=27.67.\log_3(3x-2) + \log_{0.5} = 3 \ \log_{0.5} = -1, so \log_3(3x-2) - 1 = 3 \\ \log_3(3x-2) = 4 \\ 3x - 2 = 3^4 \\ 3x - 2 = 81 \\ 3x = 83 \\ x = \frac{83}{3} = 27.67.

Step 3

Indien \( \log_2 a = b \) en \( \log_3 5 = c \), bepaal die waarde van \( \log \sqrt{0,6} \)

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Answer

Gebruik die eienskappe van logaritme:

log0,6=log(0,6)1/2=12log(0,6).\log \sqrt{0,6} = \log(0,6)^{1/2} = \frac{1}{2} \log(0,6).

Verander 0,6 in terme van logaritmes:

Step 4

Gebruik die Argand-diagram hierbo om die spanning in die vorm $V = r(\cos \theta + i \sin \theta)$ neer te skryf.

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Die Argand-diagram vertoon 'n kompleks getal wat gegee is as:

V=2(cos240°+isin240°)V = 2(\cos 240° + i \sin 240°)

Hieruit kan ons die waarde van r en ( \theta ) identifiseer. In hierdie geval is r = 2 en ( \theta = 240° ).

Step 5

Druk vervolgens, of andersins, V in gereg vorm uit.

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Die geregleerde vorm is:

V=2(cos240°+isin240°)=2(12i32)=13i.V = 2 \left( \cos 240° + i \sin 240° \right) = 2 \left( -\frac{1}{2} - i \frac{\sqrt{3}}{2} \right) = -1 - \sqrt{3}i.

Step 6

Bepaal die numeriese waardes van m en n as $m + ni = 2(6 - 4i) - ( - 7i)$.

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Eerstens, vereenvoudig die uitdrukking:

2(64i)(7i)=128i+7i=12i.2(6 - 4i) - ( -7i) = 12 - 8i + 7i = 12 - i.

Die waardes van m en n is dus ( m = 12 ) en ( n = -1 ).

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