Question

A principal of $1500 is invested at 9% interest, compounded annually. How many years will it take to accumulate $4000 or more in the account? (Use the calculator provided if necessary.)Write the smallest possible whole number answer.

Answer 1

EXPLANATION

The equation of the compounded interest is the following:

$Interest\text{ Earned=Principal\lparen1+}\frac{r}{n}\text{\rparen}^{nt}$

Where P=Principal=1500, r=rate=0.09, n = number of times interest rate is compounded = 1, t=time=unknown value

Plugging in the terms into the expression:

$4000=1500(1+\frac{0.09}{1})^t$

$\mathrm{Switch\:sides}$

$1500\left(1+\frac{0.09}{1}\right)^t=4000$

$\mathrm{Divide\:both\:sides\:by\:}1500$

$\frac{1500\left(1+\frac{0.09}{1}\right)^t}{1500}=\frac{4000}{1500}$

$\mathrm{Simplify}$

$\left(1+\frac{0.09}{1}\right)^t=\frac{8}{3}$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(\left(1+\frac{0.09}{1}\right)^t\right)=\ln \left(\frac{8}{3}\right)$

Apply log rule:

$t\ln \left(1+\frac{0.09}{1}\right)=\ln \left(\frac{8}{3}\right)$

$\mathrm{Divide\:both\:sides\:by\:}\ln \left(1.09\right)$

$\frac{t\ln \left(1.09\right)}{\ln \left(1.09\right)}=\frac{\ln \left(\frac{8}{3}\right)}{\ln \left(1.09\right)}$

Simplify:

$t=\frac{\ln \left(\frac{8}{3}\right)}{\ln \left(1.09\right)}$

Expressing as a decimal:

$t=11.38$

In conclusion, we will need 11 years to accumulate $4,000