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The book has ln(1/x), can anyone explain why not lnx? and what happened to the negative sign? Is it a typo or are they applying some logarithm rule I don't remember? It's driving me nuts. Thank you.

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    $\begingroup$ It’s $-\ln x+C$, and $-\ln x=(-1)\ln x=\ln x^{-1}=\ln\frac1x$. $\endgroup$ Commented Oct 7, 2015 at 19:11
  • $\begingroup$ See the negative sign in front of the integration. $\endgroup$
    – Empty
    Commented Oct 7, 2015 at 19:11
  • $\begingroup$ What's with the vote to close as opinion based? $\endgroup$
    – user223391
    Commented Oct 7, 2015 at 19:39

2 Answers 2

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There are two things to this.

The first thing is that you should know that $$\ln(1/x) = -\ln(x).$$ Why does this happen? It follows from the rule $\ln(a/b) = \ln(a) - \ln(b)$, and if we let $a=1$ and $b=x$, this rule gives us $\ln(1/x) = \ln(1) -\ln(x) = 0-\ln(x) = -\ln(x)$, where I use that $\ln(1) = 0$.

The second thing is that your integral has a minus sign in front of it: $$-\int \frac{1}{x}\ dx$$

Solving this, I get

$$-\int\frac{1}{x}\ dx = -\ln(x)+C = \ln(1/x)+C,$$

where I use the rule I gave (I assume that $x>0$ in all of the above).

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Here's another way to think about where the minus sign "went".

Recall that $n\ln a = \ln(a^n)$.

Therefore, $-\ln(x) = \ln(x^{-1}) = \ln(1/x) $.

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  • $\begingroup$ This may be too terse to be helpful to the original poster, although the little you said does go to the heart of what caused the confusion about the missing minus sign. $\endgroup$
    – hardmath
    Commented Oct 7, 2015 at 20:02
  • $\begingroup$ I wrote this because the previous answer explains it in another way. So, it may be possible that the poster be satisfied with this answer which is just another way of thinking. $\endgroup$ Commented Oct 8, 2015 at 12:12
  • $\begingroup$ There's a quick introduction to posting math notation using MathJax and $\LaTeX$, for future reference. I'll edit your answer to give you an idea. $\endgroup$
    – hardmath
    Commented Oct 8, 2015 at 12:17

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