Can someone go through how we get from line (i) to line (ii) ?
(i) (-3/4) ln|v-3| + (-1/4) ln|v+1| = ln x + C
(ii) (v+1)(v-3)^3 = A/ (x^4)
Trying to brush up on log rules.. Watch
- Thread Starter
- 12-12-2010 15:50
- 12-12-2010 17:10
Set C = ln c, then:
where A = 1/c^4. Are you familiar with the log rules I've used? If you really want rigorous justification for the switch from modulus signs to parentheses in line 4, you can see that the right-hand side is always positive so the left-hand side is always going to be positive (or, as in university maths courses, you can just drop the modulus and assume the bit inside a log is always positive :| ).
EDIT: Also generally speaking, the modulus bit is just a nitpicky feature of A-level maths, which makes no sense to me because the rest of A-level isn't rigorous at all. So assume you can always just drop the |.| sign whenever you switch from 'ln' to 'normal'. Or better yet just think of it as taking exp(.) of both sides.Last edited by Dream Eater; 12-12-2010 at 17:13.