Question

log(x-3)+log(x+6)=1 necesito con desarrollo porfa

Answer 1

Respuesta:

Explicación paso a paso:

log(x-3.x+6)=1

x^2+3x-18=10^1

x^2+3x=28

x(x+3)=28

x=4

Answer 2

Respuesta:

HOLA...

$log(x - 3) + log(x + 6) = 1$

$= log(x - 3) + log(x + 6) = 1si ,\: xE(3, \infty )$

$= log( {x}^{2} + 6x - 3x - 18 ) = 1$

$= {x}^{2} + 6x - 3x - 18 = 1$

$= {x}^{2} + 6x - 3x + 18 = 10 {}^1}$

$= {x}^{2} + 3x - 18 = 10$

$= {x}^{2} + 3x - 18 - 10 = 0$

$= x(x + 7) - 4(x + 7) = 0$

$= (x + 7)(x - 4) = 0$

$x + 7 = 0 \\ si, \: xE(3, \infty ) \\ x - 4 = 0$

$x = 4$

COMPROBACIÓN

$log(1) + 1 = 1 \\ = 0 + 1 = 1</p><p>$ 1=1

Answer 3

Respuesta:

HOLA...

$log(x - 3) + log(x + 6) = 1$

$= log(x - 3) + log(x + 6) = 1si, \: xE(3, \infty )$

$= log( {x}^{2} + 6x - 3x - 18 ) = 1$

$= {x}^{2} + 6x - 3x - 18 = 1$

$= {x}^{2} + 6x - 3x + 18 = 10 {}^1}$

$= {x}^{2} + 3x - 18 = 10$

$= {x}^{2} + 3x - 18 - 10 = 0$

$= x(x + 7) - 4(x + 7) = 0$

$= (x + 7)(x - 4) = 0$

$x + 7 = 0 \\ si ,\: xE(3 ,\infty ) \\ x - 4 = 0$

$x = 4$

COMPROBACIÓN

$log(1) + 1 = 1 \\ = 0 + 1 = 1$