Respuesta :
Every 10 days, triple the population. Start at 11 frogs
Day 0 -- 11 frogs
Day 10 -- 33 frogs
Day 20 -- 99 frogs
Day 30 -- 297 frogs
Etc etc...
The idea is to find which day gets us to 110 frogs. It's somewhere between day 20 and day 30
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The formula to use is
y = 11*(3)^(x/10)
where x is the number of days and y is the population. The exponent x/10 indicates that the tripling happens every 10 days. Eg: if x = 20, then x/10 = 2 indicates the second time the population has tripled.
Replace y with 110 and then use logs to isolate the exponent, leading to helping fully isolate x itself.
y = 11*(3)^(x/10)
110 = 11*(3)^(x/10)
11*(3)^(x/10) = 110
(3)^(x/10) = 110/11 ..... divide both sides by 11
(3)^(x/10) = 10
log[ (3)^(x/10) ] = log[ 10 ] .... apply logs to both sides
(x/10)*log[ 3 ] = log[ 10 ] ... use a log rule to help pull down exponent
x/10 = log(10)/log(3) .... divide both sides by log(3)
x = 10*log(10)/log(3) .... multiply both sides by 10
x = 20.95 approximately
As expected, the solution is between 20 and 30. It's much closer to x = 20 since above shows "day 20 -- 99 frogs" and 110 is close to 99.
Round x = 20.95 to the nearest whole number and we get x = 21. So on day 21 is when the population will exceed 110 frogs. If you go with x = 20 as your answer, then it will be too small. Again as the above shows, day 20 has 99 frogs instead of 110 like we want.
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