I would've thought, intuitively, it would be $$\ln(\frac{x_1}{x_2}) = \ln(x_1) - \ln(x_2)$$ But it's not. Why is this the case?
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1$\ln(x)=\ln(x/1)=\ln(1)-\ln(x)=-\ln(x)$ accoring to what you are saying. – Leo163 Oct 24 '17 at 17:16
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1Oh my, that's a good point. – sangstar Oct 24 '17 at 17:17
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What do you mean by "converse" here? – Thomas Andrews Oct 24 '17 at 17:19
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The reverse arrangement of the terms on the RHS. Leo163 illustrated the implication mathematically above. – sangstar Oct 24 '17 at 17:20
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Ah, that is merely the "reverse." Converse has a specific meaning. – Thomas Andrews Oct 24 '17 at 17:21
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Gah! I'll edit it. – sangstar Oct 24 '17 at 17:21
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1The thing you thought was right is, in fact, right. Who told you otherwise? – Wojowu Oct 24 '17 at 17:22
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1Take care : $\frac {x_1}{x_2} >0$ does not imply $x_1,x_2 >0$ – Claude Leibovici Oct 24 '17 at 17:23
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Wait, really? I was gonna say, since I was conditioned that way. In an answer for physics lecture notes, the following was obtained: $-NRT [\ln(V_2) - \ln (V_1)] = NRT \ln(\frac{V_1}{V_2})$. I reckon it's fine, since the $RHS$ is $-1$ times the $LHS$, so the fraction must switch? – sangstar Oct 24 '17 at 17:25
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There's a missing $|$ in your last comment. Anyway, note that if $0<t<1$ then $|\ln t| = -\ln t$. – David C. Ullrich Oct 24 '17 at 17:27
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@DavidC.Ullrich Right, because it's logarithmic so it starts at $(-\infty, 1)$ and becomes positive. How do I piece that together with switching the terms of the quotient? – sangstar Oct 24 '17 at 17:30