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Thread: how to avoid negative logarithms

  1. #1
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    how to avoid negative logarithms

    Hi guys,

    I am trying to solve the following inequality:

    $\frac{a. r^n}{(1 - r)} < 10^{-5}$ where a = 2 and $r = - \frac{1}{4}$


    which gives:


    $(\frac{-1}{4})^n < \frac{5}{8} \times 10^{-5}$

    If there were no minus sign attached to the 1 / 4, I could just take logs and solve for n, but with the - sign I am reduced to calculating $(- \frac{1}{4}) ^ n$ for successive values of n, until I get to my answer. Does anyone have a better way of doing this?

    thanks
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  2. #2
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    Re: how to avoid negative logarithms

    Let $n = 2k$ for some $k \in \mathbb{R}$. Then:

    $\left(-\dfrac{1}{4}\right)^n = 16^{-k}$

    So, $-k\ln 16 < \ln \left(\dfrac{5}{8}\cdot 10^{-5}\right) \Longrightarrow k > -\dfrac{\ln \left(\dfrac{5}{8}\cdot 10^{-5}\right)}{\ln 16} \Longrightarrow n>-\dfrac{\ln \left(\dfrac{5}{8}\cdot 10^{-5}\right)}{2\ln 16}$
    Thanks from s_ingram
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  3. #3
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    Re: how to avoid negative logarithms

    I'm assuming $n$ is a non-negative integer. If not, then ignore this post ...

    $\left(-\dfrac{1}{4}\right)^n = \left(-1 \cdot \dfrac{1}{4}\right)^n = (-1)^n \cdot \left(\dfrac{1}{4}\right)^n = \dfrac{(-1)^n}{4^n}$

    $\dfrac{(-1)^n}{4^n} < \dfrac{5}{8 \times 10^5}$ for all odd values of $n$

    If $n$ is even, then $(-1)^n = 1 \implies \dfrac{1}{4^n} < \dfrac{5}{8 \times 10^5} \implies 4^n > 1.6 \times 10^5 \implies n > \dfrac{\log(1.6) + 5}{\log(4)} \approx 8.644 \implies n \ge 9$
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  4. #4
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    Re: how to avoid negative logarithms

    Excellent guys!

    Once I saw SlipEternal's answer, I realised I could get to $-n log 4 < log(\frac{5}{8} \times 10^{-5})$ and it was all clear as glass. But Skeeter was too quick for me!

    Thanks to both of you!
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