How do you solve #Log(x+10) - log(x+4) = log x#?

1 Answer
sente
Dec 9, 2015

Apply properties of logarithms and solve the resulting quadratic equation to find
#x = 2# or #x = -5#

Explanation:

We will use the following properties:

  • #log(x)-log(y)=log(x/y)#

  • #e^log(x) = x#

#log(x+10) - log(x+4) = log(x)#

#=> log((x+10)/(x+4)) = log(x)#

#=> e^log((x+10)/(x+4)) = e^log(x)#

#=> (x+10)/(x+4) = x#

#=> x+10 = x(x+4) = x^2 + 4x#

#=> x^2 + 3x - 10 = 0#

#=> (x+5)(x-2) = 0#

#=> x = 2# or #x = -5#

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