If 4^{x} - 3^{x - frac{1}{2}} = 3^{x + frac{1 | vedclass

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(C) Given the equation: $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$Rearranging terms: $4^{x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} +

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$\frac{3}{2}$

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(C) Given the equation: $4^{x} - 3^{x - \frac{1}{2}} = 3^{x + \frac{1}{2}} - 2^{2x - 1}$Rearranging terms: $4^{x} + 2^{2x - 1} = 3^{x + \frac{1}{2}} + 3^{x - \frac{1}{2}}$Since $2^{2x - 1} = \frac{4^{x}}{2}$,we have: $4^{x} + \frac{4^{x}}{2} = 3^{x} \cdot \sqrt{3} + \frac{3^{x}}{\sqrt{3}}$$4^{x} \left(1 + \frac{1}{2}\right) = 3^{x} \left(\frac{3 + 1}{\sqrt{3}}\right)$$4^{x} \left(\frac{3}{2}\right) = 3^{x} \left(\frac{4}{\sqrt{3}}\right)$$\left(\frac{4}{3}\right)^{x} = \frac{4}{\sqrt{3}} \cdot \frac{2}{3} = \frac{8}{3\sqrt{3}}$$\left(\frac{4}{3}\right)^{x} = \left(\frac{2}{\sqrt{3}}\right)^{3} = \left(\left(\frac{2}{\sqrt{3}}\right)^{2}\right)^{\frac{3}{2}} = \left(\frac{4}{3}\right)^{\frac{3}{2}}$Comparing exponents,we get $x = \frac{3}{2}$.

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

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