Question

Ayuda porfa. Es pará un exámen

Answer 1

2)
$log(x + 1) = log(x - 1) + 3 \\ log(x + 1) = log(x - 1) + log( {10}^{3} ) \\ log(x + 1) = log(1000(x - 1)) \\ x + 1 = 1000(x - 1) \\ x + 1 = 1000x - 1000 \\ 1 + 1000 = 1000x - x \\ 1001 = 999x \\ x = \frac{1001}{999}$
3)
$2 log(x) - log(x + 6) = 0 \\ log( \frac{ {x}^{2} }{x + 6} ) = log(1) \\ \frac{ {x}^{2} }{x + 6} = 1 \\ {x}^{2} = x + 6 \\ {x}^{2} - x - 6 = 0 \\ (x - 3)(x + 2)$
Sol 1: x=3
Sol2: x=-2
Pero como x>0
La solucion es x =3

4)
$log(x + 1) - log(x) = 1 \\ log( \frac{x + 1}{x} ) = log(10) \\ \frac{x + 1}{x} = 10 \\ x + 1 = 10x \\ x - 10x = - 1 \\ - 9x = - 1 \\ x = \frac{1}{9}$
5)
$log(4x - 1) - log(x - 2) = log(5) \\ log( \frac{4x - 1}{x - 2} ) = log(5) \\ \frac{4x - 1}{x - 2} = 5 \\ 4x - 1 = 5(x - 2) \\ 4x - 1 = 5x - 10 \\ - 1 + 10 = 5x - 4x \\ 9 = x$
6)
$log_{3}(x) = 4 \\ {3}^{4} = x \\ 81 = x$
7)
$log_{5}(x) + log_{5}(30) = 3 \\ log_{5}(30x) = 3 \\ {5}^{3} = 30x \\ 125 = 30x \\ x = \frac{125}{30} \\ x = \frac{25}{6}$