Many years since I have done any the problem was from a work

Many years since I have done any, the problem was from a work sheet. A Frog is in a 10 foot well, climbs 3 feet during the day and slides back 2 feet at night. At what point will the frog make it to the 10 foot level and be able to crawl out?

Solution

goes up 3 in the day and goes down 2 in the night.
at the end of the first day there\'s a gain of 3.
at the end of each succeeding day there\'s a gain of 1 (loss of 2 during the night).
end of day 1 he\'s up 3
end of day 2 he\'s up 4
end of day 3 he\'s up 5
end of day 7 he\'s up 9
end of day 8 he\'s up 10
you\'re correct.
number of days is 8.
now the formula.
let x be the number of days needed to get out of the well.
formula looks like it will be
(x-1)*(3-2) + 3 = 10
what this says is that you have x-1 days where the net gain was only 1.
that would be the loss of 2 the night before plus the gain of 3 during the day.
the + 3 means you got a net gain of 3 the first day.
so (x-1)*(3-2) + 3 = 10
solving you get
(x-1)(1) + 3 = 10
x - 1 + 3 = 10
x - 1 = 7
x = 8
i must admit i didn\'t think about it correctly the first time i looked at it.
congratulations on good logic.