Question

помогите пожалуйста, полное решение

Answer 1

Ответ:

1.

${2}^{3 - x} = 32 \\ {2}^{3 - x} = {2}^{5} \\ 3 - x = 5 \\ x = - 2$

2.

${2}^{4x - 14} = \frac{1}{64} \\ {2}^{4x - 14} = {2}^{ - 6} \\ 4x - 14 = - 6 \\ 2x = 8 \\ x = 4$

3.

${( \frac{1}{3} )}^{x - 9} = \frac{1}{81} \\ {( \frac{1}{3}) }^{x - 9} = {( \frac{1}{3}) }^{4} \\ x - 9 = 4 \\ x = 13$

4.

${( \frac{1}{2}) }^{15 - x} = 16 \\ {( \frac{1}{2} )}^{15 - x} = {( \frac{1}{2} )}^{ - 4} \\ 15 - x = - 4 \\ x = 19$

5.

${36}^{x - 7} = \frac{1}{6} \\ {6}^{2(x - 7)} = {6}^{ - 1} \\ 2x - 14 = - 1 \\ 2x = 13 \\ x = 6.5$

6.

${( \frac{1}{25}) }^{x - 1} = 5 \\ {5}^{ - 2(x - 1)} = 5 \\ - 2x + 2 = 1 \\ - 2x = - 1 \\ x = 0.5$

7.

${4}^{x - 5} = 4 \\ x - 5 = 1 \\ x = 6$

8.

${7}^{1 - 2x} = 1.96 \times {5}^{1 - 2x} \\ {7}^{1 - 2x} = \frac{49}{25} \times {5}^{1 - 2x} \\ \frac{ {7}^{1 - 2x} }{49} = \frac{ {5}^{1 - 2x} }{25} \\ {7}^{1 - 2x - 2} = {5}^{1 - 2x - 2} \\ \frac{ {7}^{ - 2 x- 1} }{ {5}^{ - 2x - 1} } = 1 \\ - 2x - 1 = 0 \\ - 2x = 1 \\ x = - 0.5$

9.

${2}^{3 - x} = 0.4 \times {2}^{3 - x} \\ {2}^{ 3 - x} - 0.4 \times {2}^{3 - x} = 0 \\ {2}^{3 - x} (1 - 0.4) = 0 \\ {2}^{3 - x} = 0$

нет корней

10.

${4}^{x} - 9 \times {2}^{x} + 8 = 0$

пусть

${2}^{x} = t \\ {t}^{2} - 9t + 8 = 0 \\ d = 81 - 32 = 49 \\ t1 = \frac{9 + 7}{2} = 8 \\ t2 = 1 \\ \\ {2}^{x} = 8 \\ x1 = 3 \\ \\ {2}^{x} = 1 \\ x2 = 0$

Ответ: 3; 0.