Q. Solution of the equation ${4.9}^{x - 1} = 3\sqrt{(2^{2x + 1})}$ is
A) 3
B) 2
C) $\frac{3}{2}$
D) $\frac{2}{3}$
Explanation
(c)
${4.9}^{x - 1} = 3\sqrt{(2^{2x + 1})}$
$\Rightarrow \ \ \ 3^{2x - 2 - 1} = 2^{\frac{2x + 1}{2} - 2}$
$\Rightarrow \ \ 3^{2x - 3} = 2^{\frac{2x - 3}{2}}$
$\Rightarrow 2^{\frac{2x - 3}{2}} = \left( 3^{\frac{2x - 3}{2}} \right)^{2}$
$\Rightarrow 2x - 3 = 0$
$\therefore\ \ x = \frac{3}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.