SOLUTION: \log_(5) (x) + \log_(5) (x+5) = 9 I need help solving for x. I keep getting crazy numbers. Because I have base 5, I set up my equation as 5^9 = x(x+5) 1953125 = x^2 +5x

Question 885126: \log_(5) (x) + \log_(5) (x+5) = 9
I need help solving for x. I keep getting crazy numbers.
Because I have base 5, I set up my equation as
5^9 = x(x+5)
1953125 = x^2 +5x
x^2 + 5x - 1953125 = 0
I couldn't factor it and the quadratic formula didnt seem to work either. Please help! Thank you
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
log_(5) (x) + log_(5) (x+5) = 9
log_(5) [(x)(x+5)] = 9
(x)(x+5) = 5^9
x^2+5x = 1953125
x^2+5x-1953125 = 0
apply the "quadratic" formula:
x = -b+-sqrt(b^2-4ac)/(2a)
to get:
x = {1395.04, -1400.04}
throw out the negative solution (extraneous) to get
x = 1395.04
.
details of quadratic follows:

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=7812525 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 1395.04472200356, -1400.04472200356. Here's your graph:

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