Question

B) Alex invests $2000 in an account that has a 6% annual rate of growth. To the nearest year, when

Answer 1

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  

$A=P(1+\frac{r}{n})^{nt}$  

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  

$P=\ 2,000\\A=\ 3,600\\ r=6\%=6/100=0.06\\n=1$  

substitute in the formula above

$3,600=2,000(1+\frac{0.06}{1})^{t}$  

$1.8=(1.06)^{t}$  

Apply log both sides

$log(1.8)=log[(1.06)^{t}]$  

Applying property of exponents

$log(1.8)=(t)log(1.06)$  

$t=log(1.8)/log(1.06)$  

$t=10.09\ years$

Round to the nearest year

$t=10\ years$

Part 2) we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$  

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  

$P=\ 2,000\\A=\ 3,600\\ r=6\%=6/100=0.06$  

substitute in the formula above

$3,600=2,000(e)^{0.06t}$  

$1.8=[e]^{0.06t}$  

Apply ln both sides

$ln(1.8)=ln[e]^{0.06t}$  

$ln(1.8)=(0.06t)ln[e]$  

$t=ln(1.8)/0.06$  

$t=9.80\ years$

Round to the nearest year

$t=10\ years$