ASAP
Brianna invested $480 in an account paying an interest rate of 6 1/2% compounded continuously. Adam invested $480 in an account paying an interest rate of 6 3/4 % compounded quarterly. To the nearest hundredth of a year, how much longer would it take for Brianna's money to double than for Adam's money to double?

Respuesta :

Answer: it would take 7 years more for Brianna's money to double than for Adam's money to double

Step-by-step explanation:

Considering Brianna's investment,

The formula for continuously compounded interest is

A = P x e (r x t)

Where

A represents the future value of the investment after t years.

P represents the present value or initial amount invested

r represents the interest rate

t represents the time in years for which the investment was made.

e is the mathematical constant approximated as 2.7183.

From the information given,

P = $480

r = 6.5% = 6.5/100 = 0.065

A = 2 × 480 = $960

Therefore,

960 = 1800 x 2.7183^(0.065 x t)

960/1800 = 2.7183^(0.065t)

2 = 2.7183^(0.065t)

Taking ln of both sides, it becomes

Ln 2 = 0.065tln2.7183

0.693 = 0.065t

t = 0.693/0.065

t = 10.66

Approximately 17 years

Considering Adam's investment, we would apply the formula for determining compound interest which is expressed as

A = P(1+r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = 480

A = 960

r = 6.75% = 6.75/100 = 0.0675

n = 4 because it was compounded 4 times in a year.

Therefore,.

960 = 480(1 + 0.0675/4)^4 × t

960/480 = (1 + 0.016875)^4t

960/480 = (1.016875)^4t

Taking log of both sides, it becomes

Log 2 = 4t log 1.016875

0.301 = 0.0291t

t = 0.301/0.0291

t = 10.34

Approximately 10 years