The solution to the equation {9^x} - {2^{x + | vedclass

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Question

(B) Given equation: ${9^x} - {2^{x + 1/2}} = {2^{x + 3/2}} - {3^{2x - 1}}$Rewrite the terms: ${(3^2)^x} - {2^x} \cdot {2^{1/2}} = {2^x} \cdot {2^{3/2}

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${\log _{9/2}}(9/\sqrt 8 )$

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(B) Given equation: ${9^x} - {2^{x + 1/2}} = {2^{x + 3/2}} - {3^{2x - 1}}$Rewrite the terms: ${(3^2)^x} - {2^x} \cdot {2^{1/2}} = {2^x} \cdot {2^{3/2}} - \frac{{3^{2x}}}{3}$Let ${3^{2x}} = {9^x} = y$ and ${2^x} = z$.$y - z\sqrt 2 = z(2\sqrt 2) - \frac{y}{3}$$y + \frac{y}{3} = z(2\sqrt 2 + \sqrt 2)$$\frac{4y}{3} = 3z\sqrt 2$$\frac{y}{z} = \frac{9\sqrt 2}{4} = \frac{9}{2\sqrt 2} = \frac{9}{\sqrt 8}$Since $y = {9^x}$ and $z = {2^x}$,we have $\frac{{9^x}}{{2^x}} = {(\frac{9}{2})^x} = \frac{9}{\sqrt 8}$Taking $\log_{9/2}$ on both sides:$x = {\log _{9/2}}(\frac{9}{\sqrt 8})$

The solution to the equation ${9^x} - {2^{x + 1/2}} = {2^{x + 3/2}} - {3^{2x - 1}}$ is:

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