Question

Serenity invested $2,400 in an account paying an interest rate of 3.4% compounded
continuously. Assuming no deposits or withdrawals are made, how long would it
take, to the nearest tenth of a year, for the value of the account to reach $2,930?

Answer 1

Answer:

It would take 5.9 years to the nearest tenth of a year

Step-by-step explanation:

The formula of the compound continuously interest is A = P$e^{rt}$ , where

• A is the value of the account in t years

• P is the principal initially invested

• e is the base of a natural logarithm

• r is the rate of interest in decimal

∵ Serenity invested $2,400 in an account

∴ P = 2400

∵ The account paying an interest rate of 3.4%, compounded continuously

∴ r = 3.4% ⇒ divide it by 100 to change it to decimal

∴ r = 3.4 ÷ 100 = 0.034

∵ The value of the account reached to $2,930

∴ A = 2930

→ Substitute these values in the formula above to find t

∵ 2930 = 2400$e^{0.034t}$

→ Divide both sides by 2400

∴ $\frac{293}{240}$ = $e^{0.034t}$

→ Insert ㏑ in both sides

∴ ㏑($\frac{293}{240}$) = ㏑($e^{0.034t}$)

→ Remember ㏑($e^{n}$) = n

∴ ㏑($\frac{293}{240}$) = 0.034t

→ Divide both sides by 0.034 to find t

∴ 5.868637814 = t

→ Round it to the nearest tenth of a year

∴ t = 5.9 years

∴ It would take 5.9 years to the nearest tenth of a year