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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  !!!!!!!

 

 

 

cool cool cool

Nov 5, 2016
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Nov 5, 2016

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