We know that a convenient linear approximation of (1-x) is equal to -x. If I have something in the following form: $f= [1 - \alpha(Bk)^{\sigma}]\cdot C$. How can I rewrite it in log since my $x$ is elevated to $\sigma$?
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"$\textrm{(1-x) is equal to -x}$"No, I do not agree. And where is your $x$ in $f$? – callculus42 Jul 10 '21 at 11:12
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Imagine my x is a small one and the approximation is possible. https://math.stackexchange.com/questions/2320047/why-is-ln1-x-approx-x-when-x-is-small – Alessandro Jul 10 '21 at 11:14
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You mean $\ln(1-x) \approx -x$, if $x$ is small? – callculus42 Jul 10 '21 at 11:16
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Yes. In that case it would be like $-\alpha(Bk)$ if there was not the elevation to $\sigma$. How would it becomes with the elevation? thank you. – Alessandro Jul 10 '21 at 11:17
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The problem is, that you need the following expression: $f= [1 - \alpha(Bk)]^{\color{red}{\sigma}}\cdot C$, where $\sigma$ is outside the brackets. – callculus42 Jul 10 '21 at 11:23
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Yes but in this case if I apply the log to $[1 - \alpha(Ak)^{\rho}]$ it is not possible to write it as $- \alpha(Ak)^{\rho}$? It's like if $\alpha(Ak)^{\rho}$ is my x? – Alessandro Jul 10 '21 at 11:32
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Yes, you can do that. And don't forget the factor $C \rightarrow +\log (C)$ – callculus42 Jul 10 '21 at 11:35