Question

Помогите решить 2,4 и 5 уравнение с логарифмами пожалуйста

Answer 1

5.)

$\frac{1}{2}lg(x - 9) + \frac{1}{2} lg(2x - 1) = 1 \\ \frac{1}{2} \times log_{10}(x - 9) + \frac{1}{2} log_{10}(2x - 1) = 1 \\ xe < 9. + \infty > log_{10}(x - 9) + log_{10}(2x - 1) = 2 \\ log_{10}((x - 9)(2x - 1)) = 2 \\ log_{10}(2 {x}^{2} - x - 18x + 9 ) = 2 \\ 2 {x}^{2} - x - 18x + 9 = {10}^{2} \\ 2 {x}^{2} - 19x + 9 = 100 \\ 2 {x}^{2} - 19x + 9 - 100 = 0 \\ 2 {x}^{2} + 7x - 26x - 91 = 0 \\ x \times (2x + 7) - 13(2x + 7) = 0 \\ (2x + 7)(x - 13) = 0 \\ 2x + 7 = 0 \\ x - 13 = 0 \\ x = - \frac{7}{2} \\ xe < 90 + \infty > \\ x = 13$

2.)

$[tex] log_{6}( {x}^{2} + 5x) = 1 \\ log_{6}( {x}^{2} + 5x ) = 1 \\ xe < - \infty . - 5 > u < 0. + \infty > \\ {x}^{2} + 5x = 6 \\ {x}^{2} + 5x - 6 = 0 \\ {x}^{2} + 6x - x - 6 = 0 \\ x \times (x + 6) - (x + 6) \\ (x + 6)(x - 1) \\ x + 6 = 0 \\ x - 1 = 0 \\ x = - 6 \\ x = 1 \\ xe < - \infty . - 5 > u < 0. + \infty > \\ x = - 6 \\ x = 1 \\ {x}^{1} = - 6. {x}^{2} = 1$[/tex]

4.)

$[tex]2 {lg}^{2} x - 7lgx - 4 = 0 \\ 2 {lg}^{2} x - 7lgx = 4 \\ (2 {lg}^{2} - 7lg)x = 4 \\ x = \frac{4}{2 {lg}^{2} - 7lg} \\ x = \frac{4}{2 {g}^{2}l - 7gl }$[/tex]