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Math Help - simple problem

  1. #1
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    simple problem

    i finished about all of this math problem except for the ending.
    i reduced the problem to 5000^-10k

    what i need is how to find the value of k
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chinese_man_77 View Post
    i finished about all of this math problem except for the ending.
    i reduced the problem to 5000^-10k

    what i need is how to find the value of k
    we need an equation or something, you haven't given enough information
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  3. #3
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    A = (ax)(10)^-kt

    ax = 500
    t= 10
    A = 450

    i got it down to 450 = 5000^-10k
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  4. #4
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    Re:

    RE:

    Even if you set this equation equal to zero there still isn't a solution...

    -qbkr21
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  5. #5
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    its a regents question, so im pretty sure there is a solution and why would it be equal to 0 when its equal to 450
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  6. #6
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    A = (ax)(10)^-kt

    ax = 500
    t= 10
    A = 450

    i got it down to 450 = 5000^-10k
    i just dont know how to finish off the problem
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chinese_man_77 View Post
    A = (ax)(10)^-kt

    ax = 500
    t= 10
    A = 450

    i got it down to 450 = 5000^-10k

    we have 450 = 500 \cdot 10^{-10k}

    \Rightarrow \frac {450}{500} = 10^{-10k}

    \Rightarrow \log \left( \frac {9}{10} \right) = -10k ........do you understand this step?

    \Rightarrow k = \frac {\log \left( \frac {9}{10} \right)}{-10}
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  8. #8
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    nice, thanks a lot
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    we have 450 = 500 \cdot 10^{-10k}

    \Rightarrow \frac {450}{500} = 10^{-10k}

    \Rightarrow \log \left( \frac {9}{10} \right) = -10k ........do you understand this step?

    \Rightarrow k = \frac {\log \left( \frac {9}{10} \right)}{-10}

    one more quick question... i see how you got to the end of the problem, but is there any reason why you knew it was a log problem?
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by chinese_man_77 View Post
    one more quick question... i see how you got to the end of the problem, but is there any reason why you knew it was a log problem?
    the unknown was in the power. when we see that, we either need logs or the laws of exponents to solve the problem. logs was the appropriate choice here
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