SOLUTION: Solve {{{ln ((x-3)(x+2)) = 0}}} for x. The only solutions of the form {{{x = ((1+ sqrtb)/2)}}} what is b? What I did: ln((x-3)(x+2)) ln(x^2 -x-6) a= 1

Question 972538: Solve for x. The only solutions of the form what is b?
What I did: ln((x-3)(x+2))
ln(x^2 -x-6)
a= 1 b = -1 c = -6

that would come out to
Where did I go wrong?
Found 2 solutions by jim_thompson5910, Boreal:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Convert to exponential form.

Use the quadratic formula to get

So the value of b is 29
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
You need to convert the problem to a quadratic form. You had an ln(quadratic form). One thing to do is to graph it, and just see what it looks like.
The way I like to deal with ln is raise everything to the e power. Then you have a quadratic =1, and you can deal with it accordingly.
Raise everything to the e power
(x-3)(x+2)=1
x^2-x-7=0
[1 +/- sqrt (1+28)]/2
x=3.19 and -2.19
b=28
substitute in ln (0.19)(5.19) is very close to 1, and ln (1)=0
ln (-5.19)((-0.19) is the same,

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