Answer:
0.50 or about half a year longer.
Step-by-step explanation:
We can write an equation to model bot investments.
Oliver invested $970 in an account paying an interest rate of 7.5% compounded continuously.
Recall that continuous compound is given by the equation:
![A = Pe^{rt}](https://tex.z-dn.net/?f=A%20%3D%20Pe%5E%7Brt%7D)
Where <em>A</em> is the amount afterwards, <em>P</em> is the principal amount, <em>r</em> is the rate, and <em>t</em> is the time in years.
Since the initial investment is $970 at a rate of 7.5%:
![A = 970e^{0.075t}](https://tex.z-dn.net/?f=A%20%3D%20970e%5E%7B0.075t%7D)
Carson invested $970 in an account paying an interest rate of 7.375% compounded annually.
Recall that compound interest is given by the equation:
![\displaystyle A = P\left(1+\frac{r}{n}\right)^{nt}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%20%3D%20P%5Cleft%281%2B%5Cfrac%7Br%7D%7Bn%7D%5Cright%29%5E%7Bnt%7D)
Where <em>A</em> is the amount afterwards, <em>P</em> is the principal amount, <em>r</em> is the rate, <em>n</em> is the number of times compounded per year, and <em>t</em> is the time in years.
Since the initial investment is $970 at a rate of 7.375% compounded annually:
![\displaystyle A = 970\left(1+\frac{0.07375}{1}\right)^{(1)t}=970(1.07375)^t](https://tex.z-dn.net/?f=%5Cdisplaystyle%20A%20%3D%20970%5Cleft%281%2B%5Cfrac%7B0.07375%7D%7B1%7D%5Cright%29%5E%7B%281%29t%7D%3D970%281.07375%29%5Et)
When Oliver's money doubles, he will have $1,940 afterwards. Hence:
![1940= 970e^{0.075t}](https://tex.z-dn.net/?f=1940%3D%20970e%5E%7B0.075t%7D)
Solve for <em>t: </em>
![\displaystyle 2 = e^{0.075t}](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%20%3D%20e%5E%7B0.075t%7D)
Take the natural log of both sides:
![\ln\left (2\right) = \ln\left(e^{0.075t}\right)](https://tex.z-dn.net/?f=%5Cln%5Cleft%20%282%5Cright%29%20%3D%20%5Cln%5Cleft%28e%5E%7B0.075t%7D%5Cright%29)
Simplify:
![\ln(2) = 0.075t\Rightarrow \displaystyle t = \frac{\ln(2)}{0.075}\text{ years}](https://tex.z-dn.net/?f=%5Cln%282%29%20%3D%200.075t%5CRightarrow%20%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B%5Cln%282%29%7D%7B0.075%7D%5Ctext%7B%20years%7D)
When Carson's money doubles, he will have $1,940 afterwards. Hence:
![\displaystyle 1940=970(1.07375)^t](https://tex.z-dn.net/?f=%5Cdisplaystyle%201940%3D970%281.07375%29%5Et)
Solve for <em>t: </em>
![2=(1.07375)^t](https://tex.z-dn.net/?f=2%3D%281.07375%29%5Et)
Take the natural log of both sides:
![\ln(2)=\ln\left((1.07375)^t\right)](https://tex.z-dn.net/?f=%5Cln%282%29%3D%5Cln%5Cleft%28%281.07375%29%5Et%5Cright%29)
Simplify:
![\ln(2)=t\ln\left((1.07375)\right)](https://tex.z-dn.net/?f=%5Cln%282%29%3Dt%5Cln%5Cleft%28%281.07375%29%5Cright%29)
Hence:
![\displaystyle t = \frac{\ln(2)}{\ln(1.07375)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20t%20%3D%20%5Cfrac%7B%5Cln%282%29%7D%7B%5Cln%281.07375%29%7D)
Then it will take Carson's money:
![\displaystyle \Delta t = \frac{\ln(2)}{\ln(1.07375)}-\frac{\ln(2)}{0.075}=0.4991\approx 0.50](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5CDelta%20t%20%3D%20%5Cfrac%7B%5Cln%282%29%7D%7B%5Cln%281.07375%29%7D-%5Cfrac%7B%5Cln%282%29%7D%7B0.075%7D%3D0.4991%5Capprox%200.50)
About 0.50 or half a year longer to double than Oliver's money.