Question

Suppose that Upper P 0 is invested in a savings account in which interest is compounded continuously at 6.4​% per year. That​ is, the balance P grows at the rate given by the following equation. StartFraction dP Over dt EndFraction equals 0.064​P(t) ​(a) Find the function​ P(t) that satisfies the equation. Write it in terms of Upper P 0 and 0.064. ​(b) Suppose that ​$500 is invested. What is the balance after 3 ​years? ​(c) When will an investment of ​$500 double​ itself?

Answer 1

Answer:

(B) $602.2750

(C) 11.72 years

Explanation:

(A)

The compound interest formula is

$Principal \times (1+ r)^{time} = Ammount$

(B)

$Principal \times (1+ r)^{time} = Ammount$

$500 \times (1.064)^{3} = 602.2750$

(C)

$1(1.064)^{time} = 2\\\\log_{1.064}\: 2= time\\\\\frac{log 2}{log 1.064} = 11.717341521$