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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