Given Log z xy= c-1, log x yz = a-1, log y zx = b-1, show that ab + bc + ca = abc.

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Viranch Mistry · Answer

log z xy= c-1 => log z xy +1= c => log z xy + log z z = c => log z xyz = c Thus we have, c = log z xyz, a= log x xyz and b= log y xyz Now, LHS = ab + bc +ca let us first consider 'ab' ..... ab = (log x xyz).(log y xyz) = [(log e xyz)/(log e x)] . [(log e xyz)/(log e y)] = (log e xyz) 2 / logx.logy Simlarly find bc and ac and add you will get (log e xyz) 3 / logx.logy.logz And RHS = abc = (log e xyz) 3 / logx.logy.logz Therefore, LHS = RHS !! :)

Anish Nair · Answer

c -1=Log z xy c=Log z xy +1 c= Log z xy+Log z z c= Log z (xyz) --1 Similarly, b= Log y xyz a= Log x xyz put the values of a b c in given equation and get the result

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Given Log z xy= c-1, log x yz = a-1, log y zx = b-1, show that ab + bc + ca = abc.

Given Log_zxy= c-1, log_xyz = a-1, log_yzx = b-1, show that ab + bc + ca = abc.

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Given Log z xy= c-1, log x yz = a-1, log y zx = b-1, show that ab + bc + ca = abc.

Given Logzxy= c-1, logxyz = a-1, logyzx = b-1, show that ab + bc + ca = abc.

2 Answers

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Given Log_zxy= c-1, log_xyz = a-1, log_yzx = b-1, show that ab + bc + ca = abc.