Prove that:
(logax·logbx)/(logax + logbx) =logabx
`
We must assume that a, b, and x are all positive
and different from 1, otherwise the above is not defined.
`
We need two rules of logarithms
(1) logmn = 1/lognm, when m and n are positive but neither is 1
(2) logmp + logmq = logm(pq), when m,p,q are positive and m is not 1
`
Use (1) to replace every term in left hand side
1 1
------- · -------
logxa logxb
-----------------------
1 1
------- + --------
logxa logxb
Multiply top and bottom by logxa·logxb
1
--------------------
logxb + logxa
1
------------
logx(ba)
` use (1)
1
------------
1
--------
logbax
logbax
logabx
Edwin