Question

Solution of the equation$9^{x} - 2^{x + \frac{1}{2}} = 2^{x + \frac{3}{2}} - 3^{2x - 1}$

• A. $\log_{9}(9/\sqrt{8})$
• B. $\log_{(9/2)}(9/\sqrt{8})$
• C. $\log_{e}(9/\sqrt{8})$
• D. None of these

Answer 1

Answer: (B)

$\log_{(9/2)}(9/\sqrt{8})$

Answer 2

$9^{x} - 2^{x + (1/2)} = 2^{x + (3/2)} - 3^{2x - 1}$

⇒ $3^{2x} + \frac{1}{3}.3^{2x} = 2.2^{x + \frac{1}{2}} + 2^{x + \frac{1}{2} - 2}$

⇒ $4.3^{2x - 1} = 3.2^{x + \frac{1}{2}}$ ⇒ $3^{2x - 2} = 2^{x + \frac{1}{2} - 2}$

⇒ $3^{2x - 2} = 2^{x - \frac{3}{2}}$ ⇒ $\left( \frac{9}{2} \right)^{x - 1} = 2^{- 1/2}$

⇒ $(x - 1)\log_{9/2}9/2 = - \frac{1}{2}\log_{9/2}2$

⇒ $x - 1 = - \frac{1}{2}\log_{9/2}2$

⇒ $x = 1 - \log_{9/2}\sqrt{2} = \log_{9/2}9/2 - \log_{9/2}\sqrt{2}$

⇒ $x = \log_{9/2}(9/2\sqrt{2})$; $\therefore$ $x = \log_{9/2}(9/\sqrt{8})$.