Question

Первая строчка, логарифмы. Помогите пожалуйста

Answer 1

${9}^{x} - {2}^{x + \frac{1}{2} } = {2}^{x + \frac{7}{2} } - {3}^{2x - 1} \\ {3}^{2x} - {2}^{x} \times {2}^{ \frac{1}{2} } = {2}^{x} \times {2}^{ \frac{7}{2} } - {3}^{2x} \times {3}^{ - 1} \\ {3}^{2x} + {3}^{2x} \times {3}^{ - 1} = {2}^{2x} \times {2}^{ \frac{7}{2} } + {2}^{x} \times {2}^{ \frac{1}{2} } \\ {3}^{2x} (1 + \frac{1}{3} ) = {2}^{x} ( {2}^{ \frac{7}{2} } + {2}^{ \frac{1}{2} } ) \\ {9}^{x} \times \frac{4}{3} = {2}^{x} \times {2}^{ \frac{1}{2} } ( {2}^{3} + 1) \\ {9}^{x} \times \frac{4}{3} = {2}^{x} \times 9 \sqrt{2} \\ \frac{ {9}^{x} }{ {2}^{x} } = \frac{9 \sqrt{2} }{ \frac{4}{3} } \\ ( \frac{9}{2} ) {}^{x} = \frac{27 \sqrt{2} }{4} \\ ( \frac{9}{2} ) {}^{x} = \frac{27}{2 \sqrt{2} } \\ ( \frac{9}{2} ) {}^{x} = \frac{ {3}^{3} }{ {2}^{ \frac{3}{2} } } \\ ( \frac{9}{2} ) {}^{x} = \frac{9 \sqrt{9} }{ {2}^{ \frac{3}{2} } } \\ ( \frac{9}{2} ) {}^{x} = ( \frac{9}{2} ) {}^{ \frac{3}{2} } \\ x = \frac{3}{2}$

Ответ: 1.5