The value of x in the given equation 4^x - 3^ | vedclass

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(B) Given 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}$$2^{2x}(1 + 1

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$3/2$

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(B) Given 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}$$2^{2x}(1 + 1/2) = 3^{x - 1/2}(3 + 1)$$2^{2x} \cdot (3/2) = 3^{x - 1/2} \cdot 4$$2^{2x} \cdot 3 \cdot 2^{-1} = 3^{x - 1/2} \cdot 2^2$$2^{2x - 1} \cdot 3 = 3^{x - 1/2} \cdot 2^2$$2^{2x - 3} = 3^{x - 1/2 - 1} = 3^{x - 3/2}$Taking logarithm on both sides:$(2x - 3) \log 2 = (x - 3/2) \log 3$$2x \log 2 - 3 \log 2 = x \log 3 - (3/2) \log 3$$x(2 \log 2 - \log 3) = 3 \log 2 - (3/2) \log 3$$x \log(4/3) = \log(2^3) - \log(3^{3/2})$$x \log(4/3) = \log(8 / (3\sqrt{3}))$Since $8 / (3\sqrt{3}) = (4/3)^{3/2}$,we have:$x \log(4/3) = (3/2) \log(4/3)$Therefore,$x = 3/2$.

The value of $x$ in the given equation $4^x - 3^{x - 1/2} = 3^{x + 1/2} - 2^{2x - 1}$ is

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