SOLUTION: 2log(x^2+4x)=1 Professor says there are two answers?

Question 301264: 2log(x^2+4x)=1
Professor says there are two answers?
Found 2 solutions by stanbon, nerdybill:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
2log(x^2+4x)=1
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log(x^2+4x) = 1/2
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x^2+4x = 10^(1/2)
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x^2 + 4x - sqrt(10) = 0
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Use Quadratic Formula:
x = [-4 +- sqrt(16 - 4(-sqrt(10))]/2
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Those are the two solutions.
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Cheers,
Stan H.
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
2log(x^2+4x)=1
log(x^2+4x)^2=1
(x^2+4x)^2=10^1
(x^2+4x)^2=10
x^2+4x=sqrt(10)
x^2+4x-sqrt(10) = 0
x^2+4x-3.162 = 0
Using the quadratic formula we get:
x={0.676, -4.676}
.
Details of quadratic:

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=28.6491106406735 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.676243198995259, -4.67624319899526. Here's your graph:

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