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Thread: Need help in solving logarithmic equation

  1. #1
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    Need help in solving logarithmic equation

    A ball is made of rubber, when dropped on a concrete floor, bounces up to 80% of the height from which it was dropped. The ball is dropped from 2m. After how many bounces will the ball not rise over 0.5 m?

    This is what I did

    .5 = 2(.8) ^n

    .25 = .8^n

    ln .25 = n ln .8

    n = 6.2

    Now, what would be the answer 6 drops or 7 drops
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  2. #2
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    Hello, daunder!

    A ball is made of rubber, when dropped on a concrete floor,
    bounces up to 80% of the height from which it was dropped.
    The ball is dropped from 2m.
    After how many bounces will the ball not rise over 0.5 m?
    The height after $\displaystyle n$ bounces is: .$\displaystyle h \;=\;2(0.8)^n$

    We want an inequality: .$\displaystyle 2(0.8)^n \:\leq \:0.5$

    Then: .$\displaystyle (0.8)^n \:\leq \:0.25 $

    Take logs: .$\displaystyle \ln(0.8^n) \:\leq \:\ln(0.25) \quad\Rightarrow\quad n\!\cdot\!\ln(0.8) \:\leq \:\ln(0.25) $

    Note: .$\displaystyle \ln(0.8)$ is a negative quantity.

    Divide by $\displaystyle \ln(0.8)\!:\;\;n \:\geq \:\frac{\ln(0.25)}{\ln(0.8)} \:=\: 6.2125...$

    Therefore: .$\displaystyle n = 7$

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  3. #3
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    Thanks for explaining
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  4. #4
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    Quote Originally Posted by daunder View Post
    A ball is made of rubber, when dropped on a concrete floor, bounces up to 80% of the height from which it was dropped. The ball is dropped from 2m. After how many bounces will the ball not rise over 0.5 m?

    This is what I did

    .5 = 2(.8) ^n

    .25 = .8^n

    ln .25 = n ln .8

    n = 6.2

    Now, what would be the answer 6 drops or 7 drops
    $\displaystyle 2(.8)^n < .5$

    $\displaystyle (.8)^n < .25$

    $\displaystyle n\ln(.8) < \ln(.25)$

    $\displaystyle n > \frac{\ln(.25)}{\ln(.8)}$

    $\displaystyle n > 6.2$

    that would be 7
    Last edited by skeeter; Apr 2nd 2009 at 05:08 PM. Reason: ... too slow tonite.
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