Help ASAP

log _a+b m  + log _a-b m -2log_a+b m*log_a-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)]  -2log_a+b √[ (a -b)(a + b)] *log_a-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[log_a+b√(a -b) + log_a+b √(a +b)] [log_a-b √ (a -b) + log_a-b √(a + b) ]  =

1/2 +  log _a+b √(a - b)  +1/2 + log _a-b √(a + b) - 2 [ log_a+b√(a -b) + 1/2] [ 1/2 + log_a-b √(a + b) ]  =

1 + log _a+b √(a - b) + log _a-b √(a + b) - 2[ 1/4 + (1/2)log_a-b √(a + b) + (1/2) log_a+b√(a -b) +  log_a+b√(a -b)log_a-b √(a + b)]  =

1 + log _a+b √(a - b) + log _a-b √(a + b) - (1/2) - log_a-b √(a + b) - log_a+b√(a -b) -2 log_a+b√(a -b)log_a-b √(a + b)  =

1/2 + log _a+b √(a - b) +  log _a-b √(a + b) - log_a-b √(a + b) - log_a+b√(a -b) - 2 log_a+b√(a -b)log_a-b √(a + b)  =

1/2 - 2 log_a+b√(a -b)log_a-b √(a + b)  

1/2 - 2 [(1/2) log_a+b (a - b) (1/2)log_a-b (a + b) ]  =

1/2 - 2[1/4] [ log_a+b (a + b) * log_a-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  !!!!!!!

Like so:

I've a feeling there must be a simpler way!!

Alan: If you feel there is a simpler way, God help the rest of us!!. Can you try Wolfram/Alpha and see what they come up with. I don't even know HOW to enter it!.

Alan: Look at W/A here and see if it makes any sense! It is simply beyond me! :

http://www.wolframalpha.com/input/?i=m%5E2%3Da%5E2-b%5E2,+Logbase(a%2Bb)(m)%2BLogbase(a-b)(m)-2Logbase(a%2Bb)(m)*Logbase(a-b)(m)

CPhill: Congrats. simply brilliant that would get the same result, manually, as Alan did using computer software!!. Look at W/A result and see if you can make any sense of it.

Thanks...but after that one....maybe I should have used a "crutch".....LOL!!!!!

Dear Blarney Banker,

Is it possible for you to compliment someone without slighting someone else, or slighting the complimented himself?  This is a rhetorical question. I don’t expect an answer. The answer is obvious, anyway: It’s “no.” 

Sir Alan didn’t use software to solve this. He used software (Texmaker) to compose and display the equations that he manually solved. He used Mathcad to test the solution. Sir CPhill also used software to compose and display his solution. His software is a web browser. Neither used software to SOLVE it. They both solved it “manually.” I know this is antithetical to your thought process, but it’s a fact.

Here are two recent examples of your fecal-based blarney:

http://web2.0calc.com/questions/magic-number-just-for-fun

http://web2.0calc.com/questions/there-are-240-animals-in-a-farm-40-goats-are-added-and-the-fraction-av-goats-is-tripled-what-is-the-new-amount-of-goats

Last year, I was taking an Abnormal Psychology class. Had I known of you then, you would have provided excellent anecdotal examples of parasitic human baboon social behavior.

Fret not! I decided to immortalize you in a work of art, because I finally found a use for a quarter-size marbleized plaster form originally titled “The Stud,” that I ruined two years ago. This statue was to be a complement to the marble variations of Venus de Milo that most of the male students were creating.  A careless swipe with a chisel changed him into “The Eunuch.”  That would have been OK, but while changing his eyes from those of lustful awe to those of fateful pain, I made them too beady and crooked; that seemed to end his life.

However, after getting to know your character, I reconsidered this naked catastrophe for a bit. I realize all he needs is an Irish puint symbol in each eye and a leprechaun’s pot of gold at his feet. His birthday suit already lacks pockets, so, with these additions, it’s ready to become the immortalization of the Banker from Killarney. This will be a masterpiece of trolling art, depicting the leader of worthless gnomes who wander homeless, void of active intelligence, leaving piles of blarney in the Kingdom of Camelot.   

GA

Please Help!!

Formula:

\(\begin{array}{|rcll|} \hline x^{\log_x(y)} &=& y \quad &|\quad \log_y() \\ \log_y(x^{\log_x(y)}) &=& \log_y(y) \quad &|\quad \log_y(y)=1 \\ \log_y(x^{\log_x(y)}) &=& 1 \\ \mathbf{\log_x(y)\cdot \log_y(x) } &\mathbf{=} & \mathbf{1} \\ \hline \end{array} \)

We substitute:

\(\begin{array}{|rcll|} \hline \log_{a+b}(m) &=& \dfrac{1 }{\log_{m}(a+b)} \\ \log_{a-b}(m) &=& \dfrac{1 }{\log_{m}(a-b)} \\ \hline \end{array}\)

So:

\(\begin{array}{|lcll|} \hline \log_{a+b}(m) + \log_{a-b}(m) -2\cdot \log_{a+b}(m) \cdot \log_{a-b}(m) \\ & \log_{a+b}(m) = \dfrac{1 }{\log_{m}(a+b)} \\ & \log_{a-b}(m) = \dfrac{1 }{\log_{m}(a-b)} \\ = \dfrac{1}{\log_{m}(a+b)} + \dfrac{1}{\log_{m}(a-b)} -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{\log_{m}(a-b)+\log_{m}(a+b)}{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{\log_{m}[(a-b)\cdot(a+b)] }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{\log_{m}[(a^2-b^2)] }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \quad &|\quad a^2-b^2 = m^2\\\\ = \dfrac{\log_{m}(m^2) }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = \dfrac{2\cdot \log_{m}(m) }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \quad &|\quad \log_{m}(m) = 1\\\\ = \dfrac{2 }{\log_{m}(a+b)\cdot \log_{m}(a+b) } -2\cdot \dfrac{1}{\log_{m}(a+b)} \cdot \dfrac{1}{\log_{m}(a-b)} \\\\ = 0 \\ \hline \end{array} \)