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

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> 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      Log On


   

Question 972538: Solve ln+%28%28x-3%29%28x%2B2%29%29+=+0 for x. The only solutions of the form x+=+%28%281%2B+sqrtb%29%2F2%29 what is b?
What I did: ln((x-3)(x+2))
ln(x^2 -x-6)
a= 1 b = -1 c = -6
x+=+%28-%281%29+%2B-+sqrt%28%28-1%29%5E2+-+4%28a%29%28-6%29%29%29%2F2%281%29
that would come out to x+=+%281+%2B-+sqrt%2825%29%29%2F2
Where did I go wrong?
Found 2 solutions by jim_thompson5910, Boreal:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
ln+%28%28x-3%29%28x%2B2%29%29+=+0


%28x-3%29%28x%2B2%29+=+e%5E0 Convert to exponential form.


%28x-3%29%28x%2B2%29+=+1


x%5E2-x-6+=+1


x%5E2-x-6-1=0


x%5E2-x-7=0


Use the quadratic formula to get


x+=+%281+%2B-+sqrt%2829%29%29%2F2


So the value of b is 29

Answer by Boreal(15235) About Me  (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,
+graph+%28300%2C300%2C-10%2C10%2C-10%2C10%2Cln%28%28x-3%29%28x%2B2%29%29%29