Answer:
He invest for 2 years.
Step-by-step explanation:
Given : Suppose you have $1,950 in your savings account at the end of a certain period of time. You invested $1,700 at a 6.88% simple annual interest rate.
To find : How long, in years, did you invest your money?
Solution :
Applying simple interest formula,
![A=P(1+r)^t](https://tex.z-dn.net/?f=A%3DP%281%2Br%29%5Et)
Where, A is the amount A=$1950
P is the principal P=$1700
r is the interest rate r=6.88%=0.0688
t is the time
Substitute the values in the formula,
![1950=1700(1+0.0688)^t](https://tex.z-dn.net/?f=1950%3D1700%281%2B0.0688%29%5Et)
![\frac{1950}{1700}=(1.0688)^t](https://tex.z-dn.net/?f=%5Cfrac%7B1950%7D%7B1700%7D%3D%281.0688%29%5Et)
![1.147=(1.0688)^t](https://tex.z-dn.net/?f=1.147%3D%281.0688%29%5Et)
Taking log both side,
![\log(1.147)=\log ((1.0688)^t)](https://tex.z-dn.net/?f=%5Clog%281.147%29%3D%5Clog%20%28%281.0688%29%5Et%29)
Applying logarithmic formula, ![\log a^x=x\log a](https://tex.z-dn.net/?f=%5Clog%20a%5Ex%3Dx%5Clog%20a)
![\log(1.147)=t\log (1.0688)](https://tex.z-dn.net/?f=%5Clog%281.147%29%3Dt%5Clog%20%281.0688%29)
![t=\frac{\log(1.147)}{\log (1.0688)}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Clog%281.147%29%7D%7B%5Clog%20%281.0688%29%7D)
![t=2.06](https://tex.z-dn.net/?f=t%3D2.06)
Approximately, He invest for 2 years.