Question

6000 dollars is invested in a bank account at an interest rate of 7 percent per year, compounded continuously. Meanwhile, 33000 dollars is invested in a bank account at an interest rate of 3 percent compounded annually. To the nearest year, When will the two accounts have the same balance?

Answer 1

Answer:

The answer is "4".

Explanation:

Formula:

compounded continuously:

$A(t)= pe^{rt}$

compounded annually:

$A(t)=P(1+r)^t$

If compounded continuously = compounded annually

$\Rightarrow pe^{rt}=P(1+r)^t\\\\\Rightarrow 6000e^{0.07t}=33000(1+0.03)^t\\\\\Rightarrow \frac{e^{0.07t}}{(1+0.03)^t}=\frac{33000}{6000} \\\\\Rightarrow \frac{e^{0.07t}}{(103)^t}=\frac{33}{6} \\\\ \Rightarrow (\frac{e^{0.07}}{(103)})^{t}=\frac{33}{6} \\\\ \ Taking \log on \ both \ sides:$

$\Rightarrow \log \frac{e^{0.07}}{(103)}^{t}=\log (\frac{33}{6}) \\\\\Rightarrow t \times 0.17 =0.74\\\\ \Rightarrow t= \frac{0.74}{0.17}\\\\\Rightarrow t= 4.35\ \ \ or \ \ \ 4$