13.48x - 200 < 256.12
1 answer:
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To model this situation, we are going to use the compound interest formula: 
where
is the final amount after 
years
is the initial deposit
is the interest rate in decimal form
is the number of times the interest is compounded per year
is the time in years
For account A:
We know for our problem that
and 
. Since the interest is compounded monthly, it is compounded 12 times per year; therefore, 
. Lets replace those values in our formula:
For account B:
, 
, . Lest replace those values in our formula:
Since we want to find the time,, <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>
Now that we have our equation, we just need to solve for:
![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)
![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)
We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>
where
For account A:
We know for our problem that
For account B:
Since we want to find the time,
</span>
Now that we have our equation, we just need to solve for
We can conclude that after 17.47 years <span>the sum of the balance in both accounts will be at least 5000.</span>