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As I understand it, the first-order integrated rate law is:

$$\ln[\ce{A}] = \ln[\ce{A}]_0 - k t$$

However, I'm also told that this can be expressed as a ratio of $[\ce{A}]_0$ and $[\ce{A}]$, as follows

$$\ln \left(\frac{[\ce{A}]_0}{[\ce{A}]\,\,}\right) = kt$$

How did they arrive at this expression of the integrated rate law? I don't see a way to arrive at it algebraically, unless I'm greatly missing something.

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closed as off-topic by Mithoron, andselisk, Geoff Hutchison, bon, airhuff Jan 25 '18 at 18:28

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    $\begingroup$ This is just using logarithm rule $\log(a/b) = \log(a)-\log(b)$. $\endgroup$ – King Tut Jan 25 '18 at 13:29
  • $\begingroup$ Interesting, I was unaware of that rule. Thank you for the clarification and sorry for wasting your time. $\endgroup$ – anonymous2 Jan 26 '18 at 0:30
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To get to ln([A0]/[A])=kt from ln[A]=−kt+ln[A]0 just requires some understandings of log properties and a few log rules.

First rearrange ln[A]=−kt+ln[A]0

to

kt = ln[A]0 - ln[A]

From there, the log rule log(a/b)=log(a)−log(b) means that you can state this as

kt = ln([A]0/[A])

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