Question

Помогите решить, пожалуйста ​

Answer 1

Ответ:

$1)9 {}^{ - x} = 27 \\ 3 {}^{ - 2x} = 3 {}^{3} \\ - 2x = 3 \\ x = - \frac{3}{2}$

$2) \frac{1}{8} \times \sqrt{2 {}^{x - 1} } = 4 {}^{ - 1.25} \\ \frac{1}{2 {}^{3} } \times 2 {}^{ \frac{x - 1}{2} } = 2 {}^{ - 2.5} \\ 2 {}^{ \frac{x - 7}{2} } = 2 {}^{ - 2.5} \\ \frac{x - 7}{2} = - 2.5 \\ x - 7 = - 5 \\ x = - 5 + 7 \\ x = 2$

$3) {5}^{x + 1} - 3 \times {5}^{x - 2} = 122 \\ (5 {}^{3} - 3) \times {5}^{x - 2} = 122 \\ (125 - 3) \times {5}^{x - 2} = 122 \\ 122 \times {5}^{x - 2} = 122 \\ {5}^{x - 2} = 1 \\ {5}^{x - 2} = 5 {}^{0} \\ x - 2 = 0 \\ x = 2$

$4)9 {}^{x } - 2 \times 3 {}^{x} = 63 \\ (3 {}^{2} ) {}^{x} - 2 \times 3 {}^{x} = 63 \\ (3 {}^{x} ) {}^{2} - 2 \times 3 {}^{x} = 63 \\ t {}^{2} - 2t = 63 \\ t = - 7 \\ t = 9 \\ {3}^{x} = - 7 \\ {3}^{x} = 9 \\ x = 2$

$5)(sin \frac{5\pi}{6} ) {}^{3x - 1} = \sqrt{8} \\ ( \frac{1}{2}) {}^{3x - 4} = 2 {}^{ \frac{3}{2} } \\ 2 {}^{ - 3 + 4} = 2 {}^{ \frac{3}{2} } \\ - 3x + 4 = \frac{3}{2} \\ - 3x = \frac{3}{2} - 4 \\ - 3x = - \frac{5}{2} \\ x = \frac{5}{6}$

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