Suppose that $20,000 is invested at 9% interest, compounded annually. After time t, in years, it grows to the amount A given by the f

Question

Suppose that $20,000 is invested at 9% interest, compounded annually. After time t, in years, it grows to the amount A given by the function A(t) = $20,000(1.09)
a) Find the amount of time after which there will be $200,000 in the account.
b) Find the doubling time.
After what amount of time will there be $200,000 in the account?

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Ngọc Diệp 4 days 2021-07-19T06:26:11+00:00 1 Answers 0 views 0

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    2021-07-19T06:28:08+00:00

    Answer:

    Years = {log(total) -log(Principal)} ÷ log(1 + rate)

    Years = [log (200,000) -log (20,000)] / log (1.09)

    Years = [5.3010299957 -4.3010299957] /  0.037426497941

    Years = 1 / 0.037426497941

    Years = 26.7190374471

    *************************************************

    Finding Doubling Time : (We’ll calculate $100 becoming $200)

    Years = {log(total) -log(Principal)} ÷ log(1 + rate)

    Years = log (200) -log (100) / log(1.09)

    Years =  (2.3010299957 -2) / 0.037426497941

    Years = .3010299957 / 0.037426497941

    Years 8.0432317278

    It takes 8.0432317278 years for money to double at 9% per year.

    Source:

    http://www.1728.org/compint2.htm

    Step-by-step explanation:

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )