SOLUTION: Hi, I have a math packet due on the 27th, and I've been stuck on this problem the whole summer. We don't get credit if we don't show our work, so I was hoping you could help me fi

Question 153388: Hi, I have a math packet due on the 27th, and I've been stuck on this problem the whole summer. We don't get credit if we don't show our work, so I was hoping you could help me find the formula to solve this:
"Snap your fingers. Wait one second and snap them again. Then wait 2 seconds and snap them again. Following this pattern, the next snap will come 4 seconds after that, than 8, 16 and so on. Each time, the number of seconds between snaps is doubled. If you follow this pattern carefully, how many times times will you snap your fingers in one year? Explain your reasoning."
I've already tried y=2^2x+1, 2^31,536,000 (because there are 31,536,000 seconds in a year), I've tried everything Icould think of, but nothing made ssensse. Please help!!!!
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
this problem is somewhat analogous to counting in binary

decimal-binary __ 0-0, 1-1, 2-10, 3-11, 4-100, 5-101, 6-110, 7-111, 8-1000

if there are 2 snaps after the starting snap, it takes 3 sec
__ if there are 3 snaps after the starting snap, it takes 7 sec

if n is the number of snaps after the starting snap, then 2^n-1 is the number of seconds

2^n-1=31536000 __ adding 1 __ 2^n=31536001 __ taking log __ n[log(2)]=log(31536001)

n=[log(31536001)]/[log(2)] __ n=24.91 (approx)

this is rounded down (truncated) to 24 because 25 would take more than a year
__ remembering to add in the starting snap, gives 25 snaps in a year
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