There are currently 11 frogs in a (large) pond. the frog population grows exponentially, tripling every 10 days. how long will it take (in days) for there to be 110 frogs in the pond?

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

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.

75

step-by-step explanation:

you do x-29=46 (the number of cantaloupes sold)

in order to find x, you have to find what minus 29 equals 46, meaning you'd have to add 29 to 46.

29+46=75

if you replace x with 75, the equation looks like:

75-29=46

x=75