Помогите, пожалуйста, с алгеброй. показательные ур-ния. именно два что обведены очень...

$9^{x} - 2^{x+0,5} = 2^{x+3,5} - 3^{2x-1}\\3^{2x} - 2^{x+0,5} = 2^{x+3,5} - 3^{2x-1}\\3^{2x} + 3^{2x-1} = 2^{x+3,5} + 2^{x+0,5}\\3^{2x-1}(3 + 1) = 2^{x}(2^{3,5}+2^{0,5})\\3^{2x-1} \cdot 4 = 2^{x}(8\sqrt{2} +\sqrt{2})\\4 \cdot 3^{2x-1} = 9\sqrt{2} \cdot 2^{x}\\4 \cdot \dfrac{9^{x}}{3} = 9\sqrt{2} \cdot 2^{x}\\\bigg(\dfrac{9}{2} \bigg)^{x} = \dfrac{27\sqrt{2}}{4}\\\bigg(\dfrac{9}{2} \bigg)^{x} = \dfrac{9}{2} \cdot \dfrac{3\sqrt{2}}{2} \\\bigg(\dfrac{9}{2} \bigg)^{x-1} = \dfrac{3}{\sqrt{2}} \\$

$\bigg(\dfrac{9}{2} \bigg)^{x-1} = \bigg(\dfrac{9}{2} \bigg)^{0,5}\\x - 1 = 0,5\\x = 1,5$

0\\4t^{2} - t - 18 = 0\\t_{1} = -2 < 0\\t_{2} = \dfrac{9}{4}\\\bigg(\dfrac{2}{3} \bigg)^{x} = \dfrac{9}{4}\\\bigg(\dfrac{2}{3} \bigg)^{x} = \bigg(\dfrac{2}{3} \bigg)^{-2}\\x = -2" alt="2^{2x+2} - 6^{x} - 2 \cdot 3^{2x+2} = 0\\2^{2} \cdot 2^{2x} - (2 \cdot 3)^{x} - 2 \cdot 3^{2} \cdot 3^{2x} = 0\\4 \cdot 2^{x} - 2^{x} \cdot 3^{x} - 18 \cdot 3^{2x}=0 \ \ \ | : 3^{2x} \\4 \cdot \bigg(\dfrac{2}{3} \bigg)^{2x} - \bigg(\dfrac{2}{3} \bigg)^{x} - 18 = 0\\\bigg(\dfrac{2}{3} \bigg)^{x} = t, \ t>0\\4t^{2} - t - 18 = 0\\t_{1} = -2 < 0\\t_{2} = \dfrac{9}{4}\\\bigg(\dfrac{2}{3} \bigg)^{x} = \dfrac{9}{4}\\\bigg(\dfrac{2}{3} \bigg)^{x} = \bigg(\dfrac{2}{3} \bigg)^{-2}\\x = -2" align="absmiddle" class="latex-formula">