what is the difference

[nanase]

Hi sir/miss,
Mind helping me in telling the difference of the two approaches? which one is correct/wrong? What I have done is I tried substituting a value, and they result in different answers. but I don't know which one is right /wrong. Algebraically both steps are allowed right?

 

[Harry_the_cat]

They are both correct, but the c's are different!
Note that \(\displaystyle \frac{1}{6}ln(6x-2) = \frac{1}{6}ln(2(3x-1)) = \frac{1}{6}ln(3x-1) + \frac{1}{6}ln(2)\).
The last term is a constant.

 

[nanase]

They are both correct, but the c's are different!
Note that \(\displaystyle \frac{1}{6}ln(6x-2) = \frac{1}{6}ln(2(3x-1)) = \frac{1}{6}ln(3x-1) + \frac{1}{6}ln(2)\).
The last term is a constant.

thankssss