We have two bank accounts, the first in the bank A and the

Question

We have two bank accounts, the first in the bank A and the second in the bank B. In the bank A we obtain a continuous compound rate of 2% and in the bank B we are paid a periodic compound interes rate of 7% in 3 periods (compounding frequency) . We initially deposit 12 000 euros in the bank A and 8000 in B. How long time is it necessary until the money in both accounts is exactly the same? Remark: To obtain a correct answer it is necessary to work with at least 5 decimal digits.

Profes A. · Accepted Answer

To find out when the money in both bank accounts will be exactly the same, we'll set up equations for the future values of both accounts and solve for time.

Bank A:

The formula for continuous compounding is given by:

A (t) = P_{0} \cdot e^{r t}

where: - $P_{0} = 12, 000$ euros (initial deposit), - $r = 0.02$ (2% continuous compound rate), - $t$ is the time in years, - $A (t)$ is the amount at time $t$ .

Plugging in the values for bank A, we have:

A (t) = 12000 \cdot e^{0.02 t}

Bank B:

The formula for periodic compounding is given by:

B (t) = P_{0} \cdot {(1 + \frac{r}{n})}^{n t}

where: - $P_{0} = 8, 000$ euros (initial deposit), - $r = 0.07$ (7% periodic compound interest rate), - $n = 3$ (compounding frequency), - $t$ is the time in years, - $B (t)$ is the amount at time $t$ .

Plugging in the values for bank B, we have:

B (t) = 8000 \cdot {(1 + \frac{0.07}{3})}^{3 t}

B (t) = 8000 \cdot {(1.0233333)}^{3 t}

Setting them equal:

To find the time when both accounts have the same amount of money, we set:

12000 \cdot e^{0.02 t} = 8000 \cdot {(1.0233333)}^{3 t}

Solving for $t$ :

Taking the natural logarithm on both sides:

\ln (12000) + 0.02 t = \ln (8000) + 3 t \cdot \ln (1.0233333)

Solve for $t$ :

0.02 t - 3 t \cdot \ln (1.0233333) = \ln (8000) - \ln (12000)

t (0.02 - 3 \cdot \ln (1.0233333)) = \ln (\frac{8000}{12000})

t = \frac{\ln (\frac{4}{6})}{0.02 - 3 \cdot \ln (1.0233333)}

Calculating the values:

\ln (\frac{4}{6}) = \ln (\frac{2}{3}) = - 0.4054651

3 \cdot \ln (1.0233333) \approx 3 \cdot 0.0230666 = 0.0691998

t = \frac{- 0.4054651}{0.02 - 0.0691998}

t = \frac{- 0.4054651}{- 0.0491998}

t \approx 8.238056

Therefore, it will take approximately 8.23806 years for the amount of money in both bank accounts to be exactly the same, correct to five decimal places.

We have two bank accounts, the first in the bank A and the

Bank A:

Bank B:

Setting them equal:

Solving for $t$ :

Envie suas dúvidas pelo App. Baixe agora

Prefere professores para aulas particulares ou resolução de atividades?

We have two bank accounts, the first in the bank A and the

Bank A:

Bank B:

Setting them equal:

Solving for t:

Envie suas dúvidas pelo App. Baixe agora

Prefere professores para aulas particulares ou resolução de atividades?

Solving for $t$ :