Short Questions (Question 5 – 8)

Posted on April 21, 2020 by user

Question 5

Solve the equation,

$\log_{9} (x - 2) = \log_{3} 2$

Solution:

$\begin{array}{l} \log_{9} (x - 2) = \log_{3} 2 \\ \log_{a} b = \frac{\log_{c} b}{\log_{c} a} \Rightarrow \frac{\log_{3} (x - 2)}{\log_{3} 9} = \log_{3} 2 \\ \frac{\log_{3} (x - 2)}{2} = \log_{3} 2 \\ \log_{3} (x - 2) = 2 \log_{3} 2 \\ \log_{3} (x - 2) = \log_{3} 2^{2} \\ x - 2 = 4 \\ x = 6 \end{array}$

Question 6

Solve the equation,

$\log_{9} (2 x + 12) = \log_{3} (x + 2)$

Solution:

$\begin{array}{l} \log_{9} (2 x + 12) = \log_{3} (x + 2) \\ \frac{\log_{3} (2 x + 12)}{\log_{3} 9} = \log_{3} (x + 2) \\ \log_{3} (2 x + 12) = 2 \log_{3} (x + 2) \\ \log_{3} (2 x + 12) = \log_{3} {(x + 2)}^{2} \\ 2 x + 12 = x^{2} + 4 x + 4 \\ x^{2} + 2 x - 8 = 0 \\ (x + 4) (x - 2) = 0 \\ x = - 4 (not accepted) \\ x = 2 \end{array}$

Question 7

Solve the equation,

$\log_{4} x = \frac{3}{2} \log_{2} 3$

Solution:

$\begin{array}{l} \log_{4} x = \frac{3}{2} \log_{2} 3 \\ \frac{\log_{2} x}{\log_{2} 4} = \frac{3}{2} \log_{2} 3 \\ \frac{\log_{2} x}{2} = \frac{3}{2} \log_{2} 3 \\ \log_{2} x = 2 \times \frac{3}{2} \log_{2} 3 \\ \log_{2} x = 3 \log_{2} 3 \\ \log_{2} x = \log_{2} 3^{3} \\ x = 27 \end{array}$

Question 8

Solve the equation,

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

Solution:

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