2006-07-27 02:40:23 · 8 answers · asked by aqua456 2 in Science & Mathematics ➔ Mathematics
n · 2^(n - 1) = 80
n · 2^(n - 1) = 2^4 · 5
n · 2^(n - 1) = 2^4 · 5
log2 [n · 2^(n - 1)] = log2 [2^4 · 5]
log2 [n] + log2 [2^(n - 1)] = log2 [2^4] + log2 [5]
log2 [n] + (n - 1) = 4 + log2 [5]
log2 [n] - log2 [5] = 5 - n
log2 [n / 5] = 5 - n ... or ... 2^(5 - n) = n / 5
I can see n = 5 because log2 [1] = 0, but I'm still working on a way to get to it, algebracially.
:(
2006-07-27 03:03:57 · answer #1 · answered by Anonymous · 1⤊ 0⤋
n = 5
2006-07-27 02:44:35 · answer #2 · answered by Forest_aude 3 · 0⤊ 0⤋
I dont understand your formula. Even if I did you dont solve a formula. You can isolate a variable, or express a relationship in terms of a set of variables, or such.
Your equation looks like
n* (2^(n-1))=80
but it could also be
(n_2)^(n-1)=80
or
n) 2^(n-1)=80
So which is it, and what are you looking for?
2006-07-27 02:46:37 · answer #3 · answered by Curly 6 · 0⤊ 0⤋
You can change it to log form.
(n-1)log 2=log 80
log 80/log 2 = n-1
Put log 80 and log 2 in your calculator..
Then solve for n.
2006-07-27 02:45:35 · answer #4 · answered by knifelvr 4 · 0⤊ 0⤋
5*(2)^(5-1)
=5*(2^4)
=5*16
=80
Its very simple only.If u know send more questions like this
2006-07-27 02:53:46 · answer #5 · answered by cute girl 1 · 0⤊ 0⤋
Do you mean n*2^(n-1)=80?
In that case
log(n*2^(n-1))=log(80)
log(n)+log(2^(n-1))=log(80)
log(n)+(n-1)*log(2)=log(80)
log(n)+n*log(2)=log(80)+log(2)
log(n)+n*log(2)=log(160)
I don't know how to continue...
2006-07-27 02:50:27 · answer #6 · answered by Magnus L 1 · 0⤊ 0⤋
Jelli is correct
2006-07-27 02:48:49 · answer #7 · answered by bala 1 · 0⤊ 0⤋
do it yourself! It is easy
2006-07-27 02:43:52 · answer #8 · answered by AKL 3 · 0⤊ 0⤋