Bab 5 Indeks dan Logaritma

Posted on October 15, 2016 by user

5.6 Indeks dan Logaritma, SPM Praktis (Soalan Panjang)

Soalan 3:

Diberi bahawa p= 3^r dan q = 3^t, ungkapkan yang berikut dalam sebutan rdan/ atau t.

(a) \log_{3} \frac{p q^{2}}{27},

(b) log₉p – log₂₇ q.

Penyelesaian:

(a)

Diberi p = 3^r, log₃ p = r

q= 3^t, log₃ q =t

\log_{3} \frac{p q^{2}}{27}

= log₃ pq² – log₃27

= log₃ p + log₃ q² – log₃ 3³

= r + 2 log₃ q – 3 log₃ 3

= r + 2 log₃ q – 3(1)

= r + 2t – 3

(b)

log₉ p– log₂₇ q

\begin{array}{l} = \frac{\log_{3} p}{\log_{3} 9} - \frac{\log_{3} q}{\log_{3} 27} \\ = \frac{r}{\log_{3} 3^{2}} - \frac{t}{\log_{3} 3^{3}} \\ = \frac{r}{2 \log_{3} 3} - \frac{t}{3 \log_{3} 3} \\ = \frac{r}{2} - \frac{t}{3} \end{array}

Soalan 4:

(a) Permudahkan:

log₂(2x + 1) – 5 log₄ x²+ 4 log₂ x

(b) Seterusnya, selesaikan persamaan:

log₂(2x + 1) – 5 log₄ x²+ 4 log₂ x= 4

Penyelesaian:

(a)

log₂ (2x + 1) – 5 log₄ x²+ 4 log₂ x

\begin{array}{l} = \log_{2} (2 x + 1) - \frac{5 \log_{2} x^{2}}{\log_{2} 4} + 4 \log_{2} x \\ = \log_{2} (2 x + 1) - \frac{5}{2} \log_{2} x^{2} + \log_{2} x^{4} \\ = \log_{2} (2 x + 1) - \log_{2} {(x^{2})}^{(\frac{5}{2})} + \log_{2} x^{4} \end{array}

\begin{array}{l} = \log_{2} (2 x + 1) - \log_{2} x^{5} + \log_{2} x^{4} \\ = \log_{2} \frac{(2 x + 1) (x^{4})}{x^{5}} \\ = \log_{2} \frac{2 x + 1}{x} \end{array}

(b)

log₂ (2x + 1) – 5 log₄ x²+ 4 log₂ x= 4

\begin{array}{l} \log_{2} \frac{2 x + 1}{x} = 4 \\ \frac{2 x + 1}{x} = 2^{4} \\ \frac{2 x + 1}{x} = 16 \\ 2 x + 1 = 16 x \\ 14 x = 1 \\ x = \frac{1}{14} \end{array}