How to show that 1/(2a) ln |(x - a)/(x + a)| + C is equal to 1/(2a) ln |(x + a)/(x - a)| + K?

2 Answers
Jun 28, 2018

See below.

Explanation:

1/(2a)ln|(x-a)/(x+a)|+C=1/(2a)ln(|(x+a)/(x-a)|^-1)+C
1/(2a)ln|(x-a)/(x+a)|+C=-1/(2a)ln|(x+a)/(x-a)|+C
1/(2a)ln|(x-a)/(x+a)|+C=1/(2a)ln|(x+a)/(x-a)|-1/aln|(x+a)/(x-a)|+C
Let K=-1/aln|(x+a)/(x-a)|+C so that the new constant K "absorbs" the extra -1/aln|(x+a)/(x-a)| along with C.
:.1/(2a)ln|(x-a)/(x+a)|+C=1/(2a)ln|(x+a)/(x-a)|+K

Jun 28, 2018

You can't snow it because it's not correct!

Explanation:

This is not possible. Suppose in fact that:

(1) " "1/(2a)ln abs ((x-a)/(x+a)) +C = 1/(2a)ln abs ((x+a)/(x-a))+K

Since based on the properties of logarithms:

ln abs ((x-a)/(x+a)) = -ln abs ((x+a)/(x-a))

it would follow that:

1/(2a)ln abs ((x-a)/(x+a)) +C = -1/(2a)ln abs ((x-a)/(x+a))+K

and then:

1/a ln abs ((x-a)/(x+a)) = K-C

But the function ln abs ((x-a)/(x+a)) is not constant.

graph{ln(((x-1)/(x+1))) [-40, 40, -20, 20]}

graph{ln(((x+1)/(x-1))) [-40, 40, -20, 20]}