Solve the equation: $\;$ $9^x - 2^{x + 0.5} = 2^{x + 3.5} - 3^{2x - 1}$
Given equation: $\;\;$ $9^x - 2^{x + 0.5} = 2^{x + 3.5} - 3^{2x - 1}$
i.e. $\;$ $9^x + 3^{2x} \times 3^{-1} = 2^x \times 2^{3.5} + 2^x \times 2^{0.5}$
i.e. $\;$ $9^x + \left(3^2\right)^x \times \dfrac{1}{3} = 2^x \times 2^{\frac{7}{2}} + 2^x \times 2^{\frac{1}{2}}$
i.e. $\;$ $9^x + \dfrac{9^x}{3} = 2^x \times \left(2^7\right)^{\frac{1}{2}} + 2^x \times 2^{\frac{1}{2}}$
i.e. $\;$ $9^x \left(1 + \dfrac{1}{3}\right) = 2^x \times \left(128\right)^{\frac{1}{2}} + 2^x \times 2^{\frac{1}{2}}$
i.e. $\;$ $9^x \times \dfrac{4}{3} = 2^x \times 8 \times 2^{\frac{1}{2}} + 2^x \times 2^{\frac{1}{2}}$
i.e. $\;$ $9^x \times \dfrac{4}{3} = 2^x \times 2^{\frac{1}{2}} \times 9$
i.e. $\;$ $\left(\dfrac{9}{2}\right)^x = 9 \times 9^{\frac{1}{2}} \times \dfrac{1}{2^{2 - \frac{1}{2}}}$
i.e. $\;$ $\left(\dfrac{9}{2}\right)^x = \dfrac{9^{\frac{3}{2}}}{2^{\frac{3}{2}}} = \left(\dfrac{9}{2}\right)^{\frac{3}{2}}$
$\implies$ $x = \dfrac{3}{2}$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{\dfrac{3}{2}\right\}$