# How do you arrive at the second form of the 1st order integrated rate law? [closed]

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.

## closed as off-topic by Mithoron, andselisk♦, Geoff Hutchison, bon, airhuffJan 25 '18 at 18:28

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• This is just using logarithm rule $\log(a/b) = \log(a)-\log(b)$. – King Tut Jan 25 '18 at 13:29
• Interesting, I was unaware of that rule. Thank you for the clarification and sorry for wasting your time. – anonymous2 Jan 26 '18 at 0:30