note that this isn't a summation problem, it just wants to find on what day will she get 10,000,000 dollars in 1 day
convert that the pennies to make it easier (100 pennies=1 dollar so that's 1,000,000,000 or a billion pennies) (note that 1,000,000,000=10^9)
each day is 3 times of the other
first day is 1, 2nd is 1*3, 3rd day is 1*3*3, 4th day is 1*3*3*3, etc
hmm, so after n days, the number of pennies is ![1*3^{n-1}](https://tex.z-dn.net/?f=1%2A3%5E%7Bn-1%7D)
one long way is to keep dividing by 3 to find how many times we need to do that to find n
easier way is to do this:
1,000,000,000=![3^{n-1}](https://tex.z-dn.net/?f=3%5E%7Bn-1%7D)
take ln of both sides
![ln(1,000,000,000)=ln(3^{n-1})](https://tex.z-dn.net/?f=ln%281%2C000%2C000%2C000%29%3Dln%283%5E%7Bn-1%7D%29)
![ln(10^9)=ln(3^{n-1})](https://tex.z-dn.net/?f=ln%2810%5E9%29%3Dln%283%5E%7Bn-1%7D%29)
use properties of logarithms
![9ln(10)=(n-1)ln(3)](https://tex.z-dn.net/?f=9ln%2810%29%3D%28n-1%29ln%283%29)
divide both sides by ln(3)
![\frac{9ln(10)}{ln(3)}=n-1](https://tex.z-dn.net/?f=%5Cfrac%7B9ln%2810%29%7D%7Bln%283%29%7D%3Dn-1)
add 1 to both sides
![\frac{9ln(10)}{ln(3)}+1=n](https://tex.z-dn.net/?f=%5Cfrac%7B9ln%2810%29%7D%7Bln%283%29%7D%2B1%3Dn)
using your calculator we get 19.863=n
so after 19.863 days
but we need a whole number of days so round up to 20
20 days
it will be on day 20 of the month