SOLUTION: How do you solve for X in the equation: {{{ lnX + ln(x+3) =1 }}}

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Question 381356: How do you solve for X in the equation: +lnX+%2B+ln%28x%2B3%29+=1+
Found 2 solutions by stanbon, richard1234:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
lnX + ln(x+3) =1
----
ln[x(x+3)] = 1
---
x(x+3) = 10
---
x^2+3x-10 = 0
(x-5)(x+3) = 0
---
Positive solution:
x = 5
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Cheers,
Xtan H.

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Stan, the ln is the logarithm base e, not base 10. Therefore we obtain
x%28x%2B3%29+=+e
x%5E2+%2B+3x+-+e+=+0
which yields a much uglier set of solutions.