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Simplify \(\log_{a+b}m+\log_{a-b}m -2\log_{a+b}m \cdot \log_{a-b}m\)

 

If m^2=a^2-b^2

Guest Feb 17, 2018
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log a + b m  + log a - b m   -  2log a+ b  m* log a - b m

 

Note  m^2  =  a^2 - b^2   = (a + b) (a - b)

 

And, using the change-of-base rule, we can write

 

log m / log(a + b) + log  m / log (a - b)  -  2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ] )

 

Get a common  denominator between the first two fractions and factor out log m 

 

[  log( m ) [ log (a + b) + log (a - b)]  ] /  [ log(a + b)* log(a - b)] -

2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]  

 

{ log a + log b  =  log(a * b)  }

 

log (m) [ log [ (a + b)(a - b)]  / [ log(a + b)* log(a - b)] -

2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]

 

log (m) [ log (m^2) ]  /  [ log(a + b)* log(a - b)] -

2 [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]

 

{  log a^b  =  b log a }

 

log (m) * 2 log (m) /  [ log(a + b)* log(a - b)] -

2 (  [ log m] [ log m]  /  [ log (a + b) * log (a - b) ]

 

(2 [ log (m) }^2 ( / [ log(a + b)* log(a - b)] -   (2 [ log m] ^2 ( /  [ log (a + b) * log (a - b) ] =

 

0

 

 

cool cool cool

CPhill  Feb 17, 2018
edited by CPhill  Feb 17, 2018

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