Find the solution of the equation 9^x - 2^{x | vedclass

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(B) Given equation: $9^x - 2^{x + 1/2} = 2^{x + 3/2} - 3^{2x - 1}$ Rearranging the terms: $3^{2x} + 3^{2x-1} = 2^{x + 3/2} + 2^{x + 1/2}$ Factor out c

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$\log_{(9/2)}(9/\sqrt{8})$

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(B) Given equation: $9^x - 2^{x + 1/2} = 2^{x + 3/2} - 3^{2x - 1}$ Rearranging the terms: $3^{2x} + 3^{2x-1} = 2^{x + 3/2} + 2^{x + 1/2}$ Factor out common terms: $3^{2x-1}(3 + 1) = 2^{x + 1/2}(2 + 1)$ $4 \cdot 3^{2x-1} = 3 \cdot 2^{x + 1/2}$ $3^{2x-2} = 2^{x + 1/2 - 2} = 2^{x - 3/2}$ Taking $\log_{(9/2)}$ on both sides: $(x-1) \log_{(9/2)} 9 = (x - 3/2) \log_{(9/2)} 2$ Alternatively,rewrite as: $(9/2)^{x-1} = 2^{x-1} \cdot 3^{2x-2} / 2^{x-1} = (3^2/2)^{x-1} = (9/2)^{x-1} = 2^{-1/2}$ $x - 1 = \log_{(9/2)} (2^{-1/2}) = -1/2 \log_{(9/2)} 2$ $x = 1 - \log_{(9/2)} \sqrt{2} = \log_{(9/2)} (9/2) - \log_{(9/2)} \sqrt{2} = \log_{(9/2)} (9 / (2\sqrt{2})) = \log_{(9/2)} (9/\sqrt{8})$.

Find the solution of the equation $9^x - 2^{x + 1/2} = 2^{x + 3/2} - 3^{2x - 1}$.

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