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Thread: Logarithmic word problem

  1. #1
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    Question Logarithmic word problem

    Hi,

    I hope someone can help me with the following question:

    Logarithmic word problem-screen-shot-2017-06-29-11.33.10-am.png

    Can someone please let me know whether my solution is correct? I said that in about in year 2015, the amount pests would be the same. The reason why I want to know whether I'm right is because my solution does not match up to the textbook's.

    Logarithmic word problem-img_20170629_113620-2.jpg

    I would really appreciate help!
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  2. #2
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    Re: Logarithmic word problem

    I get 15.93 years from 1997 ... the initial population of the Washington forest is 1.95 million, not 1.9 million
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  3. #3
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    Re: Logarithmic word problem

    I get a slightly different answer. I used the fact that \frac{0.97^x}{0.96^x}= \left(\frac{0.97}{0.96}\right)^x= 1.0104^x so that 1.0104^x= 1.2105. Now take the logarithm of both sides: x log(1.0104)= log(1.2105) so 0.00500x= 0.08296 and the [tex]x= \frac{0.08296}{0.00500}= 16.6, so the answer is 1996+ 17= 2013 rather than 2015. What was the answer in your text? It might be just a matter of round off error.
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    Re: Logarithmic word problem

    Quote Originally Posted by HallsofIvy View Post
    I get a slightly different answer. I used the fact that \frac{0.97^x}{0.96^x}= \left(\frac{0.97}{0.96}\right)^x= 1.0104^x so that 1.0104^x= 1.2105. Now take the logarithm of both sides: x log(1.0104)= log(1.2105) so 0.00500x= 0.08296 and the [tex]x= \frac{0.08296}{0.00500}= 16.6, so the answer is 1996+ 17= 2013 rather than 2015. What was the answer in your text? It might be just a matter of round off error.
    Yeah you're right, it was just a round off error. Thanks for your help!
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    Re: Logarithmic word problem

    230 * (1 - .04)^n = 195 * (1 - .03)^n

    Solve for n (you should get 15.93005335275....)

    Remember that a^n / b^n = (a/b)^n
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  6. #6
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    Re: Logarithmic word problem

    $2.3(0.96)^x = 1.95(0.97)^x$

    $\dfrac{2.3}{1.95} = \dfrac{0.97^x}{0.96^x} = \left(\dfrac{0.97}{0.96}\right)^x$

    $\log\left(\dfrac{2.3}{1.95}\right) = x\log\left(\dfrac{0.97}{0.96}\right)$

    $\dfrac{\log \left( \frac{2.3}{1.95} \right)}{\log \left( \frac{0.97}{0.96} \right)} = x$

    $15.93 = x$
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  7. #7
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    Re: Logarithmic word problem

    Thanks, I got it!
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