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Thread: Getting r or t out of e^rt

  1. #1
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    Getting r or t out of e^rt

    For a function such as (1+r)^n = FV/PV one can take the Log of both sides to find n

    nlog (1+r) = log(FV/PV)
    nlog (1+r) = log(FV)-log(PV)
    n = [log(FV)-log(PV)]/log (1+r)

    I need a bit of assistance in solving similar problem that has e instead of (1+r)

    e^rt = FV/PV

    What has me confuses here is the relationship between log and e as each of these is the inverse function of the other

    So do I use the same logic as I did with the earlier stated problem

    rt log(e) = log(FV/PV)
    rt log(e) = log(FV)-log(PV)
    t = [log(FV)-log(PV)]/[r log(e)]
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  2. #2
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    Re: Getting r or t out of e^rt

    Quote Originally Posted by dexteronline View Post
    For a function such as (1+r)^n = FV/PV one can take the Log of both sides to find n
    nlog (1+r) = log(FV/PV)
    nlog (1+r) = log(FV)-log(PV)
    n = [log(FV)-log(PV)]/log (1+r)
    I need a bit of assistance in solving similar problem that has e instead of (1+r)
    e^rt = FV/PV
    rt log(e) = log(FV/PV)
    rt log(e) = log(FV)-log(PV)
    t = [log(FV)-log(PV)]/[r log(e)]
    Historically $\displaystyle \log(x)$ meant $\displaystyle \log_{10}(x)$ and $\displaystyle \ln(x)$ meant $\displaystyle \log_e(x)$

    Thus in your problem $\displaystyle rt\ln(e)=rt$ because $\displaystyle \ln(e)=1$.

    Here is a warning: many today $\displaystyle \log$ to be $\displaystyle \log_e=\ln(x)$.
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  3. #3
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    Re: Getting r or t out of e^rt

    Quote Originally Posted by Plato View Post
    Historically $\displaystyle \log(x)$ meant $\displaystyle \log_{10}(x)$ and $\displaystyle \ln(x)$ meant $\displaystyle \log_e(x)$

    Thus in your problem $\displaystyle rt\ln(e)=rt$ because $\displaystyle \ln(e)=1$.

    Here is a warning: many today $\displaystyle \log$ to be $\displaystyle \log_e=\ln(x)$.
    You solved my problem, thank you
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