Question

Realizar el logaritmo log(x+√6)+log(x-√6)=1

Answer 1

Respuesta:

$\fbox{x=4}$

Explicación paso a paso:

$Log (x + \sqrt{6}) + Log (x - \sqrt{6}) = 1$

$Log [(x + \sqrt{6})*(x - \sqrt{6})] = 1$

$Log [(x + \sqrt{6})*(x - \sqrt{6})] = Log10$

$(x + \sqrt{6})*(x - \sqrt{6})] = 10$

$(x)^2 - (\sqrt{6})^2 = 10$

$(x)^2 - 6 = 10$

$(x)^2 = 10+6$

$(x)^2 = 16$

$x= \sqrt{16}$

$\fbox{x=4}$