SOLUTION: how many real solutions does the following equation have? log(4x-15)=logx+log(x+4)

Question 243826: how many real solutions does the following equation have?
log(4x-15)=logx+log(x+4)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
log(4x-15) = log(x) + log(x+4)

becomes:

log(4x-15) = log(x*(x+4)) because log(a*b) = log(a) + log(b)

In this case, a was equal to x and b was equal to (x+4)

if log(4x-15) = log(x*(x+4)) then:

4x-15 = x*(x+4) which becomes:

4x-15 = x^2 + 4x

subtract 4x from both sides of this equation to get:

x^2 + 4x - 4x = -15 which becomes:

x^2 = -15

take the square root of both sides of this equation to get:

x = +/- sqrt(-15)

since you are taking the square root of a negative number, and the result of that is not real, then the number of real solutions that this equation has is 0.

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