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Math Help - Solving for an exponential variable....did I do this right?

  1. #1
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    Solving for an exponential variable....did I do this right?

    N= N0/dn
    First, I multiplied both sides by dn : dn = N0/N
    Then I took the log of both sides: n log d = log(N0/N)
    Finally, I divided both sides by log d: n = log(N0/N)/log d
    Does this look legit?
    I was thinking of using the division log rule to manipulate it further: n = log(N0/N)-log d
    Is this correct?

    Thanks,
    ~CC
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  2. #2
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    This use of the rule is incorrect,

    \displaystyle \ln a -\ln b = \ln \left(\frac{a}{b}\right)
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  3. #3
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    Quote Originally Posted by calvin_coolidge View Post
    N= N0/dn
    First, I multiplied both sides by dn : dn = N0/N
    Then I took the log of both sides: n log d = log(N0/N)
    Finally, I divided both sides by log d: n = log(N0/N)/log d
    Does this look legit?
    I was thinking of using the division log rule to manipulate it further: n = log(N0/N)-log d
    Is this correct?

    Thanks,
    ~CC
    Up to the point where you say
    n = log(N_0/N) / log d

    I'd say everything looks right. But where is the exponential you mentioned in the title?? The solution for n is
    n = \frac{N_0}{dN}

    Why use logs?

    -Dan
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  4. #4
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    Sorry, I typed the original post using Word with super and subscripts and copied/pasted it to the page....apparently the super/subscripts didn't transfer over. N0 is N subzero...

    So, the original expression should've read:

    N = N0/d^n

    1. I first isolated d^n to give: d^n = N0/N
    2. Then I took the log of both sides: n*log d = log (N0/N)
    3. And then divided both sides by log d: n = log(N0/N)/log d

    Thanks,

    ~CC
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  5. #5
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    So is the correct use of the rule this:

    n*log d = (log N0)/(log N)....?????....
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  6. #6
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    Quote Originally Posted by calvin_coolidge View Post
    Sorry, I typed the original post using Word with super and subscripts and copied/pasted it to the page....apparently the super/subscripts didn't transfer over. N0 is N subzero...

    So, the original expression should've read:

    N = N0/d^n

    1. I first isolated d^n to give: d^n = N0/N
    2. Then I took the log of both sides: n*log d = log (N0/N)
    3. And then divided both sides by log d: n = log(N0/N)/log d

    Thanks,

    ~CC
    Correct.

    You may also write it as n = [log(N0) - log(N)]/log(d)
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