Question

ЛОГАРИФМЫ
а) ㏒²₅Х + ㏒₅Х = 2
б) ㏒₂(Х+1) = 3-㏒₂(х+3)
в) ㏑(х-2) - 1/2㏑(3х-6)=㏑2

Answer 1

1)
$log^2_5x + log_5x = 2\ log_5x = t\\ x textgreater 0\\ t^2+t-2=0\ t_1 = 1, t_2 = -2\\ log_5x=1\ x_1 = 5\\ log_5x = -2\ x_2 = frac{1}{25}$

2)
$log_2(x+1) = 3 - log_2(x+3)\\ x + 1 textgreater 0\ x+3 textgreater 0\ x textgreater -1\\ log_2(x+1) +log_2(x+3)=3\ (x+1)(x+3) = 8\ x^2 + 3x + x + 3-8=0\ x^2+4x-5=0\ x=1$

3)
$ln(x-2) - frac{1}{2}ln(3x-6) = ln2\\ x - 2 textgreater 0\ 3x-6 textgreater 0\ x textgreater 2\\ ln(x-2) - ln(sqrt{3x-6}) = ln2\\ lnfrac{x-2}{sqrt{3x-6}} = ln2\\ frac{x-2}{sqrt{3x-6}} = 2\ x - 2 = 2sqrt{3x-6}\ (x-2)^2 = 4(3x-6)\ x^2-4x+4-12x+24=0\ x^2-16x+28=0\ D = 64-28 = 36\ x_1 = 8-6=2, varnothing\ x_2=8+6=14$