# Log of 12*e3c

1. Dec 26, 2009

### Alexx1

If you have to take the log of 12*e3c
Can I do it like this: 12*ln(e3c)?
So ln e fall away, so the result is: 12*3c = 36c
Or do I have to do it like this: ln(12*e3c?

I'm not sure of it.

2. Dec 26, 2009

### Mentallic

Re: Log

Use the rule that $log(ab)=log(a)+log(b)$ and I'm sure things should fall into place

3. Dec 26, 2009

### Alexx1

Re: Log

Thanks

4. Dec 26, 2009

Re: Log

No problem