Long Questions (Question 1 – 2)

Posted on April 21, 2020 by user

5.6.1 Indices and Logarithms, SPM Practice (Long Questions)

Question 1:

(a) Find the value of

i. 2 log₂ 12 + 3 log₂5 – log₂ 15 – log₂ 150.

ii. log₈32

(b) Shows that 5ⁿ+ 5ⁿ⁺¹+ 5ⁿ⁺²can be divided by 31 for all the values of n which are positive integer.

Solution:

(a)(i)

2 log₂ 12 + 3 log₂ 5 – log₂15 – log₂ 150

= log₂ 12² + log₂ 5³– log₂ 15 – log₂ 150

= \log_{2} \frac{12^{2} \times 5^{3}}{15 \times 150}

= log₂ 8

= log₂ 2³

= 3

(a)(ii)

\begin{array}{l} \log_{8} 32 = \frac{\log_{2} 32}{\log_{2} 8} \\ = \frac{\log_{2} 2^{5}}{\log_{2} 2^{3}} = \frac{5}{3} \end{array}

(b)

5ⁿ+ 5ⁿ⁺¹+ 5ⁿ⁺²

= 5ⁿ+ (5 × 5ⁿ) + (5² × 5ⁿ)

= 5ⁿ (1 + 5 + 5²)

= 31 × 5ⁿ

Therefore, 5ⁿ+ 5ⁿ⁺¹+ 5ⁿ⁺²can be divided by 31 for all the values of n which are positive integer.

Question 2:

(a) Given log₁₀ x = 3 and log₁₀y = –2. Shows that 2xy – 10000y² = 19.

(b) Solve the equation log₃ x = log₉(x + 6).

Solution:

(a)

log₁₀x = 3 → (x = 10³)

log₁₀y = –2 → (y = 10^-2)

2xy – 10000y² = 19

Left hand side:

2xy – 10000y²

= 2 × 10³× 10^-2– 10000 (10^-2)²

= 20 – 10000 (10^-4)

= 20 – 1

= 19

= right hand side

(b)

\begin{array}{l} \log_{3} x = \log_{9} (x + 6) \\ \log_{3} x = \frac{\log_{3} (x + 6)}{\log_{3} 9} \\ \log_{3} x = \frac{\log_{3} (x + 6)}{\log_{3} 3^{2}} \\ \log_{3} x = \frac{\log_{3} (x + 6)}{2} \end{array}

2log₃ x= log₃ (x + 6)

log₃ x²= log₃ (x + 6)

x²= x + 6

x²– x – 6 = 0

(x + 2) (x – 3) = 0

x = – 2 atau 3.

log₃(– 2) not accepted (logarithm of a negative number is undefined)

Jadi, x = 3.