Log laws (possible error in answers)

[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 don't think is the same as what my answer shows me can someone point out where I am going wrong please

 

[tiny-tim]

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 don't think is the same as what my answer shows me can someone point out where I am going wrong please

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 … ?

 

[franky2727]

sorry I am 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?

 

[tiny-tim]

sorry I am 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?

Your ln|v|=-2ln|2-z|+ln|1+2z|+c is correct.

Next step:

e^ln|v| = … ?

 

[HallsofIvy]

sorry I am 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?

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?