amydamato911 amydamato911
  • 17-07-2020
  • Mathematics
contestada

What are the solutions to log (x2+8)= 1 +log (x)?

Respuesta :

surjithayer10 surjithayer10
  • 17-07-2020

Answer:

Step-by-step explanation:

log(x²+8)=1+log(x)

log(x²+8)-log(x)=1

[tex]log\frac{x^2+8}{x} =1\\\frac{x^2+8}{x} =10^1\\x^2+8=10x\\x^2-10x+8=0\\x=\frac{10 \pm \sqrt{(-10)^2-4*1*8} }{2} \\=\frac{10 \pm \sqrt{100-32} }{2} \\=\frac{10 \pm \sqrt{68} }{2} \\=\frac{10 \pm 2\sqrt{17} }{2} \\=5 \pm \sqrt{17}[/tex]

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