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Math Help - College Algebra help: exponential and logarithmic Functions

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    Exclamation College Algebra help: exponential and logarithmic Functions

    The population of the world was 3 billlion in 1960 and it increased on average at annual of 18%...
    In what year does the world population reach 5 billion ppl.
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    Quote Originally Posted by Peyton Sawyer View Post
    The population of the world was 3 billlion in 1960 and it increased on average at annual of 18%...
    In what year does the world population reach 5 billion ppl.
    Year 0: P = 3\cdot 10^9

    Year 1: P = 1.18\cdot 3 \cdot 10^9

    Year 2: P = 1.18\cdot 1.18 \cdot 3 \cdot 10^9

    Year 3: P = 1.18 \cdot 1.18 \cdot 1.18 \cdot 3 \cdot 10^9.


    Can you see that the equation for Year n would be

    P = 3\cdot 10^9 \cdot 1.18^n.


    Now let P = 5 \cdot 10^9 and solve for n.
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    Quote Originally Posted by Prove It View Post
    Year 0: P = 3\cdot 10^9

    Year 1: P = 1.18\cdot 3 \cdot 10^9

    Year 2: P = 1.18\cdot 1.18 \cdot 3 \cdot 10^9

    Year 3: P = 1.18 \cdot 1.18 \cdot 1.18 \cdot 3 \cdot 10^9.


    Can you see that the equation for Year n would be

    P = 3\cdot 10^9 \cdot 1.18^n.


    Now let P = 5 \cdot 10^9 and solve for n.
    Determine how long it takes for an investment to double its value if the interest rate is 7% compounded continuously. Show equation set up.
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    Quote Originally Posted by Peyton Sawyer View Post
    Determine how long it takes for an investment to double its value if the interest rate is 7% compounded continuously. Show equation set up.
    Please post new questions in new threads.

    Also, did you manage to finish the last question?
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    Quote Originally Posted by Prove It View Post
    Please post new questions in new threads.

    Also, did you manage to finish the last question?
    its part B to the first question.
    And im still working on finishing it up. its sort of difficult
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    For the first question:

    5\cdot 10^9 = 3\cdot 10^9 \cdot 1.18^n

    1.18^n = \frac{5}{3}

    \left(\frac{59}{50}\right)^n = \frac{5}{3}

    \ln{\left(\frac{59}{50}\right)^n} = \ln{\frac{5}{3}}

    n\ln{\frac{59}{50}} = \ln{\frac{5}{3}}

    n = \frac{\ln{\frac{5}{3}}}{\ln{\frac{59}{50}}}

     = \frac{\ln{5} - \ln{3}}{\ln{59} - \ln{50}}.



    For your second question:

    If the interest is compounding continuously then

    A = Pe^{rt}.

    Here A = 2P, r = \frac{7}{100}.

    Solve for t.
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