Answer:
![A=1500(1+0.27)^t](https://img.qammunity.org/2020/formulas/mathematics/high-school/xa485iev6t6k003boaykoo710tu9c1y7lf.png)
Rate of change per month would be 2.01%
Explanation:
Since, the exponential growth function,
![A=P(1+(r)/(n))^(nt)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/44vs2zpmywawbh2b7k4ss2gheb6z49ybcd.png)
Where,
P = principal amount,
r = annual rate,
n = number of compounding periods,
t = number of years,
Here,
P = 1500, r = 27% = 0.27, n = 1,
Thus, the number of users after t years,
![A=1500(1+0.27)^t](https://img.qammunity.org/2020/formulas/mathematics/high-school/xa485iev6t6k003boaykoo710tu9c1y7lf.png)
Let it is equivalent to number of users when it is growing at the rate of x monthly,
That is,
![1500(1+0.27)^t=1500(1+x)^(12t)](https://img.qammunity.org/2020/formulas/mathematics/high-school/s617gbt3dgj23j0kramrmn96cjvzmrfijj.png)
![1.27^t=((1+x)^(12))^t](https://img.qammunity.org/2020/formulas/mathematics/high-school/6injmluobtr246fopoy559lu0aowp6bw9l.png)
By comparing,
![(1+x)^(12)=1.27](https://img.qammunity.org/2020/formulas/mathematics/high-school/go5ttz52b3oa02petngoha6ajl6bul1n41.png)
![\implies 1 + x = 1.02012\implies x = 0.02012=2.012\%\approx 2.01\%](https://img.qammunity.org/2020/formulas/mathematics/high-school/3dw0jnu51km5u22c9f310lm1qonbo5j20z.png)