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
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