QuestionIf 8000 dollars is invested in a bank account at an interest rate of 9 per cent per year, compounded continuously. How many years will it take for your balance to reach 20000 dollars? $\square$

Studdy Solution
Solve for the unknown variable $t$.
First, divide both sides by 8000 to isolate the exponential term:
$$\frac{20000}{8000} = e^{0.09t}$$
Simplify the fraction:
$$2.5 = e^{0.09t}$$
Take the natural logarithm of both sides to solve for $t$:
$$\ln(2.5) = \ln(e^{0.09t})$$
Using the property of logarithms that $\ln(e^x) = x$, we have:
$$\ln(2.5) = 0.09t$$
Now, solve for $t$ by dividing both sides by 0.09:
$$t = \frac{\ln(2.5)}{0.09}$$
Calculate the value:
$$t \approx \frac{0.9163}{0.09}$$ $$t \approx 10.18$$
Therefore, it will take approximately $\boxed{10.18}$ years for the balance to reach $20000.