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28 July, 15:17

A country's population in 1991 was 231 million. In 1999 it was 233 million. Estimate the population in 2003 using the exponential growth formula. round your answer to the nearest million. P=Ae (kt)

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  1. 28 July, 17:06
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    Remark

    This problem is done in 2 steps. The first step determines k and the second step is your answer.

    Determining K

    P = 233 million

    A = 231 million

    k = ?

    t = 1999 - 1991 = 8 years.

    Solution

    P = Ae^ (kt)

    233 = 231 * e ^ (kt) Divide by 231

    233/231 = e^ (k8) Do the division

    1.008658 = e^ (k8) Take the log of both sides.

    ln (1.008658) = k8 * ln (e) You are in natural logs. Ln (e) = 1; kt can be brought down and made into a result that is multiplied by ln (e)

    ln (1.008658) = 8k Take the ln of 1.008 ...

    0.008621 = 8k Divide by 8

    k = 0.008621 / 8

    k = 0.0011 rounded, but a more accurate number is in the storage area of the calculator.

    Now to get the second part.

    P = ?

    A = 231

    k = 0.0011

    t = 12

    P = 231 * e^ (0.0011*12)

    P = 231 * e^ (0.012931114) using the stored value of M

    P = 231 * 1.013015083

    P = 234.006 which rounded to the closest million is 234 million

    Answer 234 million.
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