## Log vs ln

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

nathansalce 3e
Posts: 52
Joined: Thu Jul 27, 2017 3:01 am

### Log vs ln

When solving first order reactions, when is it best to use the default ln in the equation, and when is it best to convert ln into log using its 2.303 conversion?

Vincent Tse 1K
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Joined: Fri Sep 29, 2017 7:05 am
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### Re: Log vs ln

I don't really see a real need to convert to log, unless that makes the problem easier/more convenient. You should get the same answer either way if you're doing it right! Whatever's more convenient for you and your calculator/head, I suppose.

Kelly Seto 2J
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Joined: Thu Jul 27, 2017 3:00 am

### Re: Log vs ln

I would think it's always preferable to use ln since the first order reaction integrated rate law is based on the exponential decay equation that uses e

Yu Chong 2H
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

### Re: Log vs ln

Use ln, ln is used in the integrated law for a reason because it makes life easier for you. If you convert it to log then you 'll go through an extra step that you never needed to do in the first place.

Oscar Valdovinos 1I
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Joined: Fri Sep 29, 2017 7:05 am

### Re: Log vs ln

There is no need to convert ln into log, and this could cause an answer to be wrong if you ever need to get rid of ln. Since e^ln cancels out smoothly the finding the reaction constant or time is made easier. In short, stick with ln

Veronica Rasmusen 2B
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Joined: Sat Jul 22, 2017 3:01 am

### Re: Log vs ln

There aren't really any cases log will be more convenient than ln, so just stick with natural log as it will always be more simple.

Yeyang Zu 2J
Posts: 58
Joined: Fri Sep 29, 2017 7:06 am

### Re: Log vs ln

Since you can get the answer in both ways so that I don't see a real need to convert ln to log. If it makes calculation easier for you, then it is o.

Shanmitha Arun 1L
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### Re: Log vs ln

There wouldn't be any cases where you would need to use log. It's usually easiest to go by the given equation.

Shreya Ramineni 2L
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Log vs ln

There isn't a need to convert from ln to log, especially because when integrated, the first order reaction provides ln.

Tiffany 1B
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

### Re: Log vs ln

Yeah, I think you should just go with given equation. If you convert, you're adding an additional step where you could potentially introduce error.