You deposit 2000 in account A, which pays 2.25% annual interest compounded monthly. You deposit another 2000 in account b, which pays 3% annual interest compounded monthly. when is the sum of the balance in both accounts at least 5000

To model this situation, we are going to use the compound interest formula: [tex]A=P(1+ \frac{r}{n} )^{nt}[/tex]
where
[tex]A[/tex] is the final amount after [tex]t[/tex] years 
[tex]P[/tex] is the initial deposit 
[tex]r[/tex] is the interest rate in decimal form 
[tex]n[/tex] is the number of times the interest is compounded per year
[tex]t[/tex] is the time in years 

For account A: 
We know for our problem that [tex]P=2000[/tex] and [tex]r= \frac{2.25}{100} =0.0225[/tex]. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, [tex]n=12[/tex]. Lets replace those values in our formula:
[tex]A=2000(1+ \frac{0.0225}{12} )^{12t}[/tex]

For account B:
[tex]P=2000[/tex], [tex]r= \frac{3}{100} =0.03[/tex], [tex]n=12[/tex]. Lest replace those values in our formula:
[tex]A=2000(1+ \frac{0.03}{12} )^{12t}[/tex]

Since we want to find the time, [tex]t[/tex], 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:
[tex]2000(1+ \frac{0.0225}{12} )^{12t}+2000(1+ \frac{0.03}{12} )^{12t}=5000[/tex]

Now that we have our equation, we just need to solve for [tex]t[/tex]:
[tex]2000[(1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}]=5000 [/tex]
[tex](1+ \frac{0.0225}{12} )^{12t}+(1+ \frac{0.03}{12} )^{12t}= \frac{5000}
{2000} [/tex]
[tex](1.001875)^{12t}+(1.0025 )^{12t}= \frac{5} {2} [/tex]
[tex]ln(1.001875)^{12t}+ln(1.0025 )^{12t}=ln( \frac{5} {2})[/tex]
[tex]12tln(1.001875)+12tln(1.0025 )=ln( \frac{5} {2})[/tex]
[tex]t[12ln(1.001875)+12ln(1.0025 )]=ln( \frac{5} {2})[/tex]
[tex]t= \frac{ln( \frac{5}{2} )}{12ln(1.001875)+12ln(1.0025 )} [/tex]
[tex]17.47[/tex]

We can conclude that after 17.47 years the sum of the balance in both accounts will be at least 5000.