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Math Help - May I have a question?

  1. #1
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    May I have a question?

    Hi Masters,

    I calculated 0.2 = e^(-x(ln2/700)) to find x.
    But, I could not use a calculator also did not have any approximation for the value of ln.
    So, I got x = 700(ln5 / ln2).

    But the wolfram says the answer is x = 700((ln5+ 2i pi n) / ln2).
    Can I ask where 2i pi n came from?
    Was I wrong?

    Thanks a lot.
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  2. #2
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    Re: May I have a question?

    It is giving all complex solutions to the problem. Given any complex number, x= r(cos(\theta)+ i sin(\theta)= re^{i\theta}, ln(x)= ln(r)+ i\theta+ 2i n\pi, the " + 2i n\pi giving equivalent values.
    Thanks from yanirose
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  3. #3
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    Re: May I have a question?

    Hi Master

    I deeply appreciate your answer.
    I got your point.
    Then, may I ask if I was wrong if I just put 700*(ln5/ln2) in terms of the question's answer?
    Last edited by yanirose; May 3rd 2014 at 07:51 AM.
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    Re: May I have a question?

    Quote Originally Posted by yanirose View Post
    Hi Master

    I deeply appreciate your answer.
    I got your point.
    Then, may I ask if I was wrong if I just put 700*(ln5/ln2) in terms of the question's answer?
    It depends whether the question explicitly asked for a real solution or implicitly restricted the answer to the domain of the reals.
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  5. #5
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    Re: May I have a question?

    Quote Originally Posted by yanirose View Post
    Hi Masters,

    I calculated 0.2 = e^(-x(ln2/700)) to find x.
    But, I could not use a calculator also did not have any approximation for the value of ln.
    So, I got x = 700(ln5 / ln2).

    But the wolfram says the answer is x = 700((ln5+ 2i pi n) / ln2).
    Can I ask where 2i pi n came from?
    Was I wrong?

    Thanks a lot.
    Is the equation $\displaystyle \begin{align*} e^{-x\ln{ \left( \frac{2}{700} \right) }} \end{align*}$ or $\displaystyle \begin{align*} e^{\frac{-x\ln{(2)}}{700}} \end{align*}$?
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  6. #6
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    Re: May I have a question?

    Dear Masters

    The questions is the latter.
    This solution was to ask about the decay proportion, so the x is years
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  7. #7
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    Re: May I have a question?

    Well for starters, $\displaystyle \begin{align*} \frac{-x\ln{(2)}}{700} = -\frac{x}{700}\ln{(2)} = \ln{ \left( 2^{-\frac{x}{700}} \right) } \end{align*}$, so that means $\displaystyle \begin{align*} e^{-\frac{x\ln{(2)}}{700}} = e^{\ln{ \left( 2^{-\frac{x}{700}} \right) }} = 2^{-\frac{x}{700}} \end{align*}$

    I think $\displaystyle \begin{align*} 0.2 = 2^{-\frac{x}{700}} \end{align*}$ is much easier to solve
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  8. #8
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    Re: May I have a question?

    Thanks. so I got the answer for x = 700*(ln5/ln2). And I left the answer as it is because I did not have any approximation value for the log.
    I was worrying if I am wrong because the mathematica has a complex value
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