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Thread: Intersection of Exponential Equations problem.

  1. #1
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    Intersection of Exponential Equations problem.

    Problem:
    Intersection of Exponential Equations problem.-screen-shot-2017-11-12-6.44.37-pm.png

    Having the most trouble with parts 'b' and 'c'.

    Heres what I have for b.

    Intersection of Exponential Equations problem.-screen-shot-2017-11-12-6.52.24-pm.png
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  2. #2
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    Re: Intersection of Exponential Equations problem.

    Hint:
    if a^p = b, then p = log(b) / log(a)

    How did x/30 change to 87/30 ??!!
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    Re: Intersection of Exponential Equations problem.

    Since 8/60 gave me 0.1333 I thought I had to make (1/2)^x/30 = to 0.1333. When I put 87 in for x I got close to 0.1333. I'm also not sure if I'm allowed to use Logarithms in this unit, since we start learning Logarithms at the end. Heres some example questions from the same unit/section, Intersection of Exponential Equations problem.-screen-shot-2017-11-12-7.23.18-pm.png Intersection of Exponential Equations problem.-screen-shot-2017-11-12-7.23.27-pm.png
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  4. #4
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    Re: Intersection of Exponential Equations problem.

    Those examples are different: the exponent is given.
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  5. #5
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    Re: Intersection of Exponential Equations problem.

    Quote Originally Posted by Tyler010 View Post
    Since 8/60 gave me 0.1333 I thought I had to make (1/2)^x/30 = to 0.1333. When I put 87 in for x I got close to 0.1333. I'm also not sure if I'm allowed to use Logarithms in this unit, since we start learning Logarithms at the end.
    You are exactly right about what you need to do. If you can't use logarithms, you have to use trial and error, which is greatly helped by a good graph.

    In this kind of problem, you may be able to get ONLY an approximate answer. Your graph should suggest a bit less than 90 minutes. Good work with your approximation of 87. The quickest way to get an excellent approximation is to use logarithms as suggested by Denis.

    $x = \dfrac{30 * log \left ( \dfrac{8}{60} \right )}{ log \left ( \dfrac{1}{2} \right )} \approx 87.2067.$

    Now how do you check an answer?
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