Question

tyron is saving up money for a down payment on a motorcycle. he currently has $2979, but knows he can get a loan at a lower interest rate if he can put down $3830. if he invests the $2979 in an account that earns 3.8 % annually, compounded continuously, how long will it take tyron to accumulate the $3830? round your answer to two decimal places, if necessary.

Explanation:

Step1: Recall the formula for continuous compounding

The formula for continuous compounding is $A = Pe^{rt}$, where $A$ is the final amount, $P$ is the principal amount, $r$ is the annual interest rate (in decimal), and $t$ is the time in years.
We know $A = 3830$, $P = 2979$, and $r = 0.038$ (since $3.8\%=0.038$). We need to solve for $t$.

Step2: Substitute the known values into the formula

Substitute $A = 3830$, $P = 2979$, and $r = 0.038$ into $A = Pe^{rt}$:
$$3830=2979e^{0.038t}$$

Step3: Divide both sides by 2979

$$\frac{3830}{2979}=e^{0.038t}$$
Calculate $\frac{3830}{2979}\approx1.2857$
So we have $1.2857 = e^{0.038t}$

Step4: Take the natural logarithm of both sides

Take $\ln$ of both sides: $\ln(1.2857)=\ln(e^{0.038t})$
Using the property $\ln(e^x)=x$, we get:
$$\ln(1.2857)=0.038t$$

Step5: Solve for t

Divide both sides by $0.038$:
$$t=\frac{\ln(1.2857)}{0.038}$$
Calculate $\ln(1.2857)\approx0.2503$
Then $t=\frac{0.2503}{0.038}\approx6.59$

Answer:

$6.59$ years