Anonymous
Anonymous asked in Science & MathematicsMathematics · 6 months ago

exponetial?

how to solve this problems

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  • 6 months ago
    Favorite Answer

    I like to use this equation for all exponential growth and decay questions:

    a(t) = a e^(kt)

    Where "a" is the initial amount (we can use 100 here)

    e is the mathematical constant that is approximately equal to 2.71828

    t is the amount of time (in this case years)

    k is the decay (or grown if positive) constant.

    Looking at the graph, the half-life is when a(t) = 50. This is at t = 25. So knowing this information we can derive our equation to find any point:

    a(t) = a e^(kt)

    a(25) = 100 e^(k * 25)

    50 = 100 e^(25k)

    0.5 = e^(25k)

    ln(0.5) = 25k

    k = ln(0.5) / 25

    The log of a number less than 1 is negative, making this a decay function.

    Now our equation is:

    a(t) = a e^(kt)

    a(t) = 100 e^(t * ln(0.5) / 25)

    So what percentage exists after 45 years? Since it's slightly less than 2 full half-lives, we should expect to have a value near 25% (the result of two half-lives). Substitute 45 in for t and simplify:

    a(45) = 100 e^(45 * ln(0.5) / 25)

    a(45) = 100 e^(-1.24766)

    (I rounded here since I had to, but my calculator still uses the non-rounded value to reduce the chance of errors due to rounding)

    a(45) = 100(0.28717)

    a(45) = 100(0.28717)

    a(45) = 28.717

    So after 45 years, there is still 28.717% (rounded to 3DP) remaining of the initial compound.

  • Mike G
    Lv 7
    6 months ago

    i) From the graph

    50% after 25 years

    Half-life = 25 years

    ii) A = 100(0.5)^(45/25)

    = 28.717%

  • 6 months ago

    After how many years is the amount halved?

    p = 100 * 2 ^(-t/25)

    Plug in 45

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