Question

Кто понимает, помогите очень срочно ​

Answer 1

1

$log_{\frac{1}{2}}x = -5\\x = (\frac{1}{2})^{-5} = 2^5 = 32$

2.

$2x-1 = 2^6\\2x - 1 = 64\\2x = 65\\x = \frac{65}{2}$

3

$3lg{5} + lg{2} = 3lg{5*2} = 3lg10 = 3\\lgx = 3\\x = 10^3 = 1000$

4

$lgx = lg10 + lg8 = lg{(10*8)} = lg80\\x = 80$

5

$ln(\frac{x}{2}) = ln(x-5)\\\frac{x}{2} = x-5\\\\x = 2x - 10\\-x = - 10\\x = 10$

6

$log_3{(x^2 - 8x)} = log_3{9}\\x^2- 8x = 9\\x^2 - 8x - 9 = 0\\\sqrt{D} = \sqrt{(-8)^2 - 4 * (-9) * 1} = \sqrt{100} = 10\\x1 = \frac{8-10}{2} = -1\\x2 = \frac{8+10}{2} = 9$

7

$log_4{16} = 2\\2 = log_5{25}\\log_5{(2x-1)} - log_5{8} = log_5{25}\\log_5{\frac{2x-1}{8}} = log_5{25}\\\frac{2x-1}{8} = 25\\2x = 200\\x = 100$

8

$log_3{x} + log_3{(x-8)} = 2\\log_3{x} + log_3{(x-8)} = log_3{9}\\log_3{(x^2 - 8x)} = log_3{9}\\x^2 - 8x - 9 = 0\\\sqrt{D} = 10\\x1 = -1 \notin (0; +\infty)\\x2 = 9$

9

замена

$t = lgx\\t^2 + 5t + 4 = 0\\\sqrt{D} = \sqrt{25 - 16} = \sqrt{9} = 3\\t_1 = \frac{-5-3}{2} = -4\\t_2 = \frac{-5+3}{2} = -1\\lgx = -4\\lgx = -1\\x_1 = \frac{1}{10^4}\\x_2 = \frac{1}{10}$

10

$lnx(lnx - 1) = 0\\lnx = 0\\x = 1$

или

$lnx = 1\\x = e^1 = e$

11

$log_x{4} = log_4{x}\\x = 4$

12

$log_2{x} = 2^0 = 1\\x = 2$