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I can't figure out how the following approximation has been done, I would appreciate any guidance: $$y=-60+10\log_{10}\left[\frac{\left(\frac{99}{100}\right)^m}{\frac{1}{11}\left(\frac{1}{3}\right)^m+\frac{10}{11}\left(\frac{1}{6}\right)^m}\right]$$ is approximated to: $y=-49.6+4.73m$ when $m>5$ and $y=-60+7.73m$ when $m<3$. Thanks in advance.

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Let's put $y$ in logarithm and exponential form (factorizing $\left(\frac 16\right)^m=\left(\frac 13\right)^m\left(\frac 12\right)^m$) : \begin{align} y&=-60+\frac {10}{\ln(10)}\;\ln\left[\frac{e^{m\ln\left(\frac{99}{100}\right)}}{\frac{1}{11}e^{m\ln\left(\frac{1}{3}\right)}\left(1+\frac {10}{2^m}\right)}\right]\\ &=-60+\frac {10}{\ln(10)}\;\left[m\,\ln\left(\frac{99}{100}\right)-m\,\ln\left(\frac{1}{3}\right)-\ln\left(1+\frac {10}{2^m}\right)+\ln(11)\right]\\ \end{align}

At this point observe that : \begin{align} a&:=\frac {10}{\ln(10)}\left(\ln\left(\frac{99}{100}\right)+\ln(3)\right)\approx 4.73\\ b&:=-60+\frac {10}{\ln(10)}\;\ln\left(11\right)\approx -49.6\\ \end{align} If $\;m\gg 1\;$ then $\ \ln\left(1+\frac {10}{2^m}\right)\sim \frac {10}{2^m}\ $ may be neglected and you get $\ b+a\,m\ $ as wished.

For small $m$ rewrite $y$ as : \begin{align} y&=-60+\frac {10}{\ln(10)}\;\ln\left[\frac{e^{m\ln\left(\frac{99}{100}\right)}}{\frac{10}{11}e^{m\ln\left(\frac{1}{6}\right)}\left(1+\frac {2^m}{10}\right)}\right]\\ \end{align} so that using : \begin{align} c&:=\frac {10}{\ln(10)}\left(\ln\left(\frac{99}{100}\right)+\ln(6)\right)\approx 7.74\\ d&:=-60-\frac {10}{\ln(10)}\;\ln\left(\frac{10}{11}\right)\approx -59.6\\ \end{align}

you will get the approximation $\ d+c\,m\ $ (this one is not very good for 'moderate' values since valid only for $m\ll 1$.

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  • $\begingroup$ You are welcome ! $\endgroup$ Commented May 5, 2013 at 10:10

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