2006-07-27 02:00:19 · 4 answers · asked by aqua456 2 in Science & Mathematics ➔ Mathematics
n=3.77
2006-07-27 02:26:32 · answer #1 · answered by chuckynjwvu 2 · 0⤊ 1⤋
i hate to say this but all the other answers you see here are wrong this is the correct answer.
step1: 2*n^(n-1)=80
step2: n^(n-1)=40 (divide both sides by 2)
step3: [n^(n-1)]^2 = 40^2 --> n^2(n-1)= 1600
step4: n^2(n-1)=40^2
step5: 2(n-1)=2
step6: n-1=1
step7: n=2 !
2006-07-27 09:58:02 · answer #2 · answered by Carpe Diem (Seize The Day) 6 · 0⤊ 0⤋
n^(n-1) = 40 <=> (n-1)log(n) = log(40) <=> (x-0.5)log(x-0.5) = log(40) <=> ylogy = log40
you could try a Newton method
(ylogy)' = log(y) + 1
F(y) := ylogy - log40
Yn+1 = Yn - F(Yn)/F'(Yn) ; n = 1, .... until you're satisfied. take Y1 = 2 for instance.
2006-07-27 09:34:39 · answer #3 · answered by gjmb1960 7 · 0⤊ 0⤋
2*n^(n-1)=80
n^(n-1)=40
n^(n-2) = 3.77 ^ (3.77-1)
so answer is
n = 3.77
for prove 2*n^(n-1) = 2 * 3.77 ^ (3.77 - 1)
= 2 * 3.77 ^ 2.77
= 80
RHS = LHS
2006-07-27 09:34:00 · answer #4 · answered by Anonymous · 0⤊ 0⤋