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I posted this yesterday and got 0 correct answers.
I plugged the equations into Excel and found 2 empirical solutions:
n = 1.429611825
n = 8.613169459

So there are 2 roots, just like any n^2 + n + c = 0 equation.
But no one has been able to solve this mathmatically.
Best answer is waiting for you if you can show me.
I am looking for an answer I can plug into the calculator to get the exact answer of the 1.42 and 8.61 approximations that I already know.

2007-03-30 07:01:44 · 2 answers · asked by Andy C 2 in Science & Mathematics Mathematics

2 answers

This equation can only be solved algebraically through use of the Lambert W function. If you have it available, then the answer is as follows:

e^(ln n)=4 ln n
1 = 4 e^(-ln n) ln n
-1/4 = -ln n e^(-ln n)
-ln n = W(-1/4)
ln n = -W(-1/4)
n = e^(-W(-1/4))

So the question is now, does your calculator know how to compute W(x)?

2007-03-30 07:10:40 · answer #1 · answered by Pascal 7 · 3 0

Not all equations can be solved algebraically; their solutions have to be approximated. Far as I know, this is one.

2007-03-30 14:14:00 · answer #2 · answered by Philo 7 · 0 0

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