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

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(B) Given equation: $4^{x} + 2^{2x-1} = 3^{x+\frac{1}{2}} + 3^{x-\frac{1}{2}}$Since $2^{2x-1} = \frac{2^{2x}}{2} = \frac{4^{x}}{2}$,we can rewrite the

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$3/2$

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(B) Given equation: $4^{x} + 2^{2x-1} = 3^{x+\frac{1}{2}} + 3^{x-\frac{1}{2}}$Since $2^{2x-1} = \frac{2^{2x}}{2} = \frac{4^{x}}{2}$,we can rewrite the left side as:$4^{x} + \frac{4^{x}}{2} = 4^{x}(1 + \frac{1}{2}) = \frac{3}{2} \cdot 4^{x}$For the right side:$3^{x+\frac{1}{2}} + 3^{x-\frac{1}{2}} = 3^{x} \cdot 3^{\frac{1}{2}} + 3^{x} \cdot 3^{-\frac{1}{2}} = 3^{x}(\sqrt{3} + \frac{1}{\sqrt{3}}) = 3^{x}(\frac{3+1}{\sqrt{3}}) = 3^{x} \cdot \frac{4}{\sqrt{3}}$Equating both sides:$\frac{3}{2} \cdot 4^{x} = 3^{x} \cdot \frac{4}{\sqrt{3}}$$\frac{4^{x}}{3^{x}} = \frac{4}{\sqrt{3}} \cdot \frac{2}{3} = \frac{8}{3\sqrt{3}}$Since $3\sqrt{3} = 3^{1} \cdot 3^{1/2} = 3^{3/2}$ and $8 = 2^{3}$,we have:$(\frac{4}{3})^{x} = \frac{2^{3}}{3^{3/2}} = \frac{(2^{2})^{3/2}}{3^{3/2}} = (\frac{4}{3})^{3/2}$Comparing exponents,$x = 3/2$.

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

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