Find the value of x in the equation 4^x - 3^{ | vedclass

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(B) Given the equation: $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$.Rearranging the terms: $2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$.Factor

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$3/2$

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(B) Given the equation: $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$.Rearranging the terms: $2^{2x} + 2^{2x - 1} = 3^{x + 1/2} + 3^{x - 1/2}$.Factor out the common terms: $2^{2x - 1}(2 + 1) = 3^{x - 1/2}(3 + 1)$.Simplify: $2^{2x - 1} \cdot 3 = 3^{x - 1/2} \cdot 4$.$2^{2x - 1} \cdot 3 = 3^{x - 1/2} \cdot 2^2$.$2^{2x - 3} = 3^{x - 1/2 - 1} = 3^{x - 3/2}$.Taking $\log$ on both sides: $(2x - 3) \log 2 = (x - 3/2) \log 3$.$(2x - 3) \log 2 = \frac{2x - 3}{2} \log 3$.$(2x - 3) \log 2 = (2x - 3) \log \sqrt{3}$.This implies $2x - 3 = 0$ or $\log 2 = \log \sqrt{3}$ (which is impossible).Therefore,$2x - 3 = 0$,which gives $x = 3/2$.

Find the value of $x$ in the equation $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$.

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