Re: isi maths

Posted by duck on Feb 02, 2012; 2:48am
URL: http://discussion-forum.276.s1.nabble.com/isi-maths-tp7239469p7245578.html

24) |log x x_1| + |log x  x_2| + |log x / x_1| + |log x / x_2| = |log x_1 + log x_2|

LHS = |log x x_1| + |log x  x_2| + |log x / x_1| + |log x / x_2|
= |log x + log x_1| + |log x + log x_2| + |log x - log x_1| + |log x - log x_2|
= |log x + log x_1| + |log x - log x_1| + |log x + log x_2| + |log x - log x_2| (Just changing the order in which the terms are written)
= |log x + log x_1| + |log x_1 - log x| + |log x + log x_2| + |log x_2 - log x| (since |a| = |-a|)
>=   |log x + log x_1 + log x_1 - log x| +  |log x + log x_2 + log x_2 - log x| (combining first two terms together and next two terms together using |a|+|b|>=|a+b|)
= |2log x_1| +  |2 log x_2|
= 2(|log x_1| +  |log x_2|)

RHS =  |log x_1 + log x_2|
<=  |log x_1| + |log x_2|
since LHS = RHS
this implies that
|log x_1| + |log x_2| > = 2(|log x_1| +  |log x_2|)
this implies |log x_1| + |log x_2|=0
Thus, |log x_1| = |log x_2| = 0
Hence x_1 =  x_2 = 1 substituting in equation  |log x  x_1| + |log x  x_2| + |log x / x_1| + |log x / x_2| = |log x_1 + log x_2|
we get 4|log x| = 0 which gives us x = 1.
Thus there is a unique solution.