log a+b m + log a-b m -2loga+b m*loga-b m if m^2 = a^2 - b^2 .....
If m^2 = a^2 - b^2 then m^2 = (a -b)(a + b) → m = √[ (a - b)(a + b)] ...[we assume that m is positive]...so we have
log a+b √[ (a -b)(a + b)] + log a-b √[ (a -b)(a + b)] -2loga+b √[ (a -b)(a + b)] *loga-b √[ (a -b)(a + b)] =
log a+b √(a + b) + log a+b √(a - b) + log a-b √ (a - b) + log a-b √(a + b)
- 2[loga+b√(a -b) + loga+b √(a +b)] [loga-b √ (a -b) + loga-b √(a + b) ] =
1/2 + log a+b √(a - b) +1/2 + log a-b √(a + b) - 2 [ loga+b√(a -b) + 1/2] [ 1/2 + loga-b √(a + b) ] =
1 + log a+b √(a - b) + log a-b √(a + b) - 2[ 1/4 + (1/2)loga-b √(a + b) + (1/2) loga+b√(a -b) + loga+b√(a -b)loga-b √(a + b)] =
1 + log a+b √(a - b) + log a-b √(a + b) - (1/2) - loga-b √(a + b) - loga+b√(a -b) -2 loga+b√(a -b)loga-b √(a + b) =
1/2 + log a+b √(a - b) + log a-b √(a + b) - loga-b √(a + b) - loga+b√(a -b) - 2 loga+b√(a -b)loga-b √(a + b) =
1/2 - 2 loga+b√(a -b)loga-b √(a + b)
1/2 - 2 [(1/2) loga+b (a - b) (1/2)loga-b (a + b) ] =
1/2 - 2[1/4] [ loga+b (a + b) * loga-b (a + b)] = use the change of base rule in the brackets
1/2 - (1/2) [ (log (a - b) /log (a + b)] [ log(a + b) / log (a- b) ] =
1/2 - (1/2) [ log(a -b)/log(a -b)] [ log (a + b)/log (a + b) ] =
1/2 - 1/2 * [ 1] * [1 ] =
1/2 - 1/2 =
0
LOL!!!!....mine was even more complicated than Alan's !!!!!!!
