To model this situation, we are going to use the compound interest formula:
![A=P(1+ \frac{r}{n} )^{nt}](https://tex.z-dn.net/?f=A%3DP%281%2B%20%5Cfrac%7Br%7D%7Bn%7D%20%29%5E%7Bnt%7D)
where
![A](https://tex.z-dn.net/?f=A)
is the final amount after
![t](https://tex.z-dn.net/?f=t)
years
![P](https://tex.z-dn.net/?f=P)
is the initial deposit
![r](https://tex.z-dn.net/?f=r)
is the interest rate in decimal form
![n](https://tex.z-dn.net/?f=n)
is the number of times the interest is compounded per year
![t](https://tex.z-dn.net/?f=t)
is the time in years
For account A:
We know for our problem that
![P=2000](https://tex.z-dn.net/?f=P%3D2000)
and
![r= \frac{2.25}{100} =0.0225](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7B2.25%7D%7B100%7D%20%3D0.0225)
. Since the interest is compounded monthly, it is compounded 12 times per year; therefore,
![n=12](https://tex.z-dn.net/?f=n%3D12)
. Lets replace those values in our formula:
![A=2000(1+ \frac{0.0225}{12} )^{12t}](https://tex.z-dn.net/?f=A%3D2000%281%2B%20%5Cfrac%7B0.0225%7D%7B12%7D%20%29%5E%7B12t%7D)
For account B:
![P=2000](https://tex.z-dn.net/?f=P%3D2000)
,
![r= \frac{3}{100} =0.03](https://tex.z-dn.net/?f=r%3D%20%5Cfrac%7B3%7D%7B100%7D%20%3D0.03)
,
![n=12](https://tex.z-dn.net/?f=n%3D12)
. Lest replace those values in our formula:
![A=2000(1+ \frac{0.03}{12} )^{12t}](https://tex.z-dn.net/?f=A%3D2000%281%2B%20%5Cfrac%7B0.03%7D%7B12%7D%20%29%5E%7B12t%7D)
Since we want to find the time,
![t](https://tex.z-dn.net/?f=t)
, <span>when the sum of the balance in both accounts is at least 5000, we need to add both accounts and set that sum equal to 5000:
</span>
![2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000](https://tex.z-dn.net/?f=2000%281%2B%20%5Cfrac%7B0.0225%7D%7B12%7D%20%29%5E%7B12t%7D%2B2000%281%2B%20%5Cfrac%7B0.03%7D%7B12%7D%20%29%5E%7B12t%7D%3D5000)
Now that we have our equation, we just need to solve for
![t](https://tex.z-dn.net/?f=t)
:
![2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000](https://tex.z-dn.net/?f=2000%5B%281%2B%20%5Cfrac%7B0.0225%7D%7B12%7D%20%29%5E%7B12t%7D%2B%281%2B%20%5Cfrac%7B0.03%7D%7B12%7D%20%29%5E%7B12t%7D%5D%3D5000%20)
![(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000} {2000}](https://tex.z-dn.net/?f=%281%2B%20%5Cfrac%7B0.0225%7D%7B12%7D%20%29%5E%7B12t%7D%2B%281%2B%20%5Cfrac%7B0.03%7D%7B12%7D%20%29%5E%7B12t%7D%3D%20%5Cfrac%7B5000%7D%20%7B2000%7D%20)
![(1.001875)^{12t}+(1.0025 )^{12t}= \frac{5} {2}](https://tex.z-dn.net/?f=%281.001875%29%5E%7B12t%7D%2B%281.0025%20%29%5E%7B12t%7D%3D%20%5Cfrac%7B5%7D%0A%7B2%7D%20)
![ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})](https://tex.z-dn.net/?f=ln%281.001875%29%5E%7B12t%7D%2Bln%281.0025%20%29%5E%7B12t%7D%3Dln%28%20%5Cfrac%7B5%7D%20%7B2%7D%29)
![12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})](https://tex.z-dn.net/?f=12tln%281.001875%29%2B12tln%281.0025%20%29%3Dln%28%20%5Cfrac%7B5%7D%20%7B2%7D%29)
![t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})](https://tex.z-dn.net/?f=t%5B12ln%281.001875%29%2B12ln%281.0025%20%29%5D%3Dln%28%20%5Cfrac%7B5%7D%20%7B2%7D%29)
![t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7Bln%28%20%5Cfrac%7B5%7D%7B2%7D%20%29%7D%7B12ln%281.001875%29%2B12ln%281.0025%20%29%7D%20)
![17.47](https://tex.z-dn.net/?f=17.47)
We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>