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Given Logzxy= c-1, logxyz = a-1, logyzx = b-1, show that ab + bc + ca = abc.
logzxy= c-1 => logzxy +1= c => logzxy + logzz = c => logzxyz = c
Thus we have, c = logzxyz, a= logxxyz and b= logyxyz
Now, LHS = ab + bc +ca
let us first consider 'ab' ..... ab = (logxxyz).(logyxyz) = [(logexyz)/(logex)] . [(logexyz)/(logey)]
= (logexyz)2 / logx.logy
Simlarly find bc and ac and add you will get (logexyz)3 / logx.logy.logz
And RHS = abc = (logexyz)3 / logx.logy.logz
Therefore, LHS = RHS !! :)
c-1=Logzxy
c=Logzxy +1
c= Logzxy+Logzz
c= Logz(xyz) --1
Similarly,
b= Logyxyz
a= Logxxyz
put the values of a b c in given equation and get the result
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