4.5.4 Indices and Logarithms, Short Questions (Question 11 - 13)

Question 11

Given that 2 log₂ (x – y) = 3 + log₂x + log₂ y.

Prove that x² + y²– 10xy = 0.

Solution:

2 log₂ (x – y) = 3 + log₂x + log₂ y

log₂ (x– y)² = log₂ 8 + log₂ x + log₂y

log₂ (x– y)² = log₂ 8xy

(x – y)² = 8xy

x²– 2xy + y² = 8xy

x² + y² – 10xy = 0 (proven)

Question 12

Solve the equation,

$\log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6$

Solution:

$\begin{array}{l} \log_{2} 5 \sqrt{x} + \log_{4} 16 x = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{\log_{2} 4} = 6 \\ \log_{2} 5 \sqrt{x} + \frac{\log_{2} 16 x}{2} = 6 \\ 2 \log_{2} 5 \sqrt{x} + \log_{2} 16 x = 12 \\ \log_{2} {(5 \sqrt{x})}^{2} + \log_{2} 16 x = 12 \\ \log_{2} (25 x) + \log_{2} 16 x = 12 \\ \log_{2} (25 x) (16 x) = 12 \\ \log_{2} 400 x^{2} = 12 \\ 400 x^{2} = 2^{12} \\ x^{2} = 10.24 \\ x = 3.2 \end{array}$

Question 13

Solve the equation,

$\frac{2}{\log_{5} 2} = \log_{2} (2 - x)$

Solution:

$\begin{array}{l} \frac{2}{\log_{5} 2} = \log_{2} (2 - x) \\ 2 = \log_{5} 2. \log_{2} (2 - x) \\ 2 = \frac{1}{\log_{2} 5} . \log_{2} (2 - x) \\ 2 \log_{2} 5 = \log_{2} (2 - x) \\ \log_{2} 5^{2} = \log_{2} (2 - x) \\ 25 = 2 - x \\ x = - 23 \end{array}$

4.5.4 Indices and Logarithms, Short Questions (Question 11 – 13)

Leave a Comment Cancel reply