If at day 1 we get 1% better than in the day 0, we will be:
![(101)/(100)*1=1.01*1=1.01](https://img.qammunity.org/2023/formulas/mathematics/college/ytr4y5zjfxf6a08fk9rjbv6vjxjgqfmwew.png)
1.01 better on day 1 than on day 0.
If we get 1% better on day 2 than on day 1, then by day 2 we would be:
![(101)/(100)*1.01=1.01*1.01=(1.01)^2=1.0201](https://img.qammunity.org/2023/formulas/mathematics/college/w7v9my6mspr5qckxde73j6o8cpa3nmcbxs.png)
1.0201 times better on day 2 than on day 0.
After n days, we would have to multiply 1 by 1.01 n times, so by day n we would be:
![1.01^n](https://img.qammunity.org/2023/formulas/mathematics/college/8yuccsrkm0osfxmcj87ha8s3qt5p2fi2bu.png)
times better than on day 0.
Calculate 1.01^365 to find how many times better we would be one year after day 0:
![1.01^(365)=37.78343433\ldots](https://img.qammunity.org/2023/formulas/mathematics/college/evu6z0f2vc58d0peerp1foia79mc061zno.png)
Therefore, we would get 37.78 times better by day 365, which is after one year.