# Log laws (possible error in answers)

1. Aug 16, 2008

### franky2727

ive got integral[(2/(2-z))+(2/(1+2z))]dz=integral dv/v

then the next line is ln|v|=-2ln|2-z|+ln|1+2z|+c

when i equate this i get rid of the 2 as a common factor integrate the V side first and multiply through by 1/2 giving me ln|2-z|+ln|1+2z|=1/2ln|v| which i dont think is the same as what my answer shows me can someone point out where im going wrong please

2. Aug 16, 2008

### tiny-tim

Hi franky2727!

i] multiplying through by 1/2 gives you ln|2-z|+ 1/2ln|1+2z|=1/2ln|v|

ii] don't do that!

as soon as you get a log equation like ln|v|=-2ln|2-z|+ln|1+2z|+c,

just un-log it (ie do e-to-the to both sides ),

and you get … ?

3. Aug 16, 2008

### franky2727

sorry im more confused as to where the -2 has came from in there answer i thought i would get ln|v| = 2ln|2-z|+2ln|1+2z| +c or am i making serious mistakes with my log laws?

4. Aug 16, 2008

### tiny-tim

It's not the log laws that are the problem. If you are getting that, then you are integrating wrong. To integrate $\int dz/(1-z)$ let u= 1- z and to integrate $\int dz/(1+2z)$ let u= 1+ 2z. What is du in each case?