What do you think?... If a^2 - b^2 = 1, then ln |a + b| = - ln |a - b| ...Is this a correct statement?

Answer 1

yes
here's why:
a^2-b^2=1. ln|a+b|=ln| [(a+b)(a-b)] / (a-b) | = ln| (a^2-b^2) / (a - b)| =
= (since a^2-b^2=1) = ln| 1 / (a-b) | = ln | (a-b)^(-1)| = (-1)*ln|a-b|

Answer 2

If a^2 -b^2 = 1, then ln(a^2 - b^2) = ln1

And ln((a+b)(a-b)) = 0 because ln1=0

Using logarithms rule... ln(a+b) + ln(a-b) = 0 but this is true only if a>b. If we use absolute values, then it is true to all different a and b

Then we have ln|a+b| = -ln|a-b|

Answer 3

Yes, this is correct. Coz a2-b2 = (a+b).(a-b). So if you replace in the first equation, it becomes (a+b).(a-b) = 1. Now if you take log (ln) on both sides, it becomes ln [(a+b).(a-b)] = ln [1].

ln[1] is always 0 and ln [m.n] = ln [m] + ln [n]. Using this logic in the equation above, it becomes ln (a+b) + ln(a-b) = 0.

Now take one variable to the right side and it becomes

ln (a+b) = - ln(a-b)

Hence Proved

Answer 4

No that is incorrect. The correct answer is: Do you own homework and stop relying on other poeple for your personal knowledge.