If $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2 x - 1}$ , then the value of x is

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$\frac{1}{2}$

$\frac{3}{2}$

$\frac{7}{2}$

$\frac{5}{2}$

answer is D.

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Detailed Solution

$4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2 x - 1}$

$\Rightarrow 4^{x} - 3^{x} {.3}^{\frac{- 1}{2}} = 3^{x} {.3}^{\frac{1}{2}} - 2^{2 x} {.2}^{- 1}$

$\Rightarrow 4^{x} - \frac{3^{x}}{\sqrt{3}} = 3^{x} . \sqrt{3} - \frac{4^{x}}{2}$

$\Rightarrow 4^{x} + \frac{4^{x}}{2} = 3^{x} . \sqrt{3} + \frac{3^{x}}{\sqrt{3}}$

$\Rightarrow 4^{x} (1 + \frac{1}{2}) = 3^{x} (\sqrt{3} + \frac{1}{\sqrt{3}})$

$\Rightarrow \frac{3}{2} 4^{x} = \frac{4}{\sqrt{3}} {.3}^{x}$

$\Rightarrow \frac{4^{x}}{3^{x}} = \frac{2.4}{3 \sqrt{3}}$

$\Rightarrow {(\frac{4}{3})}^{x} = \frac{4^{\frac{1}{2}} {.4}^{1}}{3^{1} {.3}^{\frac{1}{2}}}$

$\Rightarrow {(\frac{4}{3})}^{x} = {(\frac{4}{3})}^{\frac{3}{2}}$

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

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