Question

Como eu resolvo esse equação de log x+log(x-2)=0 ??

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

$\mathsf{log\:x + log\:(x - 2) = 0}$

$\mathsf{log\:x(x - 2) = 0}$

$\mathsf{log\:(x^2 - 2x) = log\:1}$

$\mathsf{x^2 - 2x = 1}$

$\mathsf{x^2 - 2x - 1 = 0}$

$\mathsf{\Delta = b^2 - 4.a.c}$

$\mathsf{\Delta = (-2)^2 - 4.1.(-1)}$

$\mathsf{\Delta = 4 + 4}$

$\mathsf{\Delta = 8}$

$\mathsf{x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{2 \pm \sqrt{8}}{2} \rightarrow \begin{cases}\mathsf{x' = \dfrac{2 + 2\sqrt{2}}{2} = 1 + \sqrt{2}}\\\\\mathsf{x'' = \dfrac{2 - 2\sqrt{2}}{2} = 1 - \sqrt{2}}\end{cases}}$

$\boxed{\boxed{\mathsf{S = \{\:1 + \sqrt{2}\:\}}}}$

Answer 2

Resposta:

log x+log(x-2)=0     ...cuidado x-2>0  ==>x>2

log x*(x-2) =0

x*(x-2)=10⁰

x²-2x=1

x²-2x-1=0

x'=[2+√(4+4)]/2= 1+√2

x''=[2-√(4+4)]/2= 1-√2    <  2  , ñ serve  

x=1+√2  é a resposta