Answer:
part 1) 10 years
part 2) 10 years
Step-by-step explanation:
The correct question is:
Part 1) Alex invests $2000 in an account that has a 6% annual rate of growth compounded annually. To the nearest year, when will the investment be worth $3600?
Part 2) Alex invests $2000 in an account that has a 6% annual rate of growth compounded continuously. To the nearest year, when will the investment be worth $3600?
Part 1) we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
[tex] P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06\\n=1[/tex]
substitute in the formula above
[tex]3,600=2,000(1+\frac{0.06}{1})^{t}[/tex]
[tex]1.8=(1.06)^{t}[/tex]
Apply log both sides
[tex]log(1.8)=log[(1.06)^{t}][/tex]
Applying property of exponents
[tex]log(1.8)=(t)log(1.06)[/tex]
[tex]t=log(1.8)/log(1.06)[/tex]
[tex]t=10.09\ years[/tex]
Round to the nearest year
[tex]t=10\ years[/tex]
Part 2) we know that
The formula to calculate continuously compounded interest is equal to
[tex]A=P(e)^{rt}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
e is the mathematical constant number
we have
[tex] P=\$2,000\\A=\$3,600\\ r=6\%=6/100=0.06[/tex]
substitute in the formula above
[tex]3,600=2,000(e)^{0.06t}[/tex]
[tex]1.8=[e]^{0.06t}[/tex]
Apply ln both sides
[tex]ln(1.8)=ln[e]^{0.06t}[/tex]
[tex]ln(1.8)=(0.06t)ln[e][/tex]
[tex]t=ln(1.8)/0.06[/tex]
[tex]t=9.80\ years[/tex]
Round to the nearest year
[tex]t=10\ years[/tex]
