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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.


Grade:8

2 Answers

Viranch Mistry
32 Points
10 years ago

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 !! :)

Anish Nair
19 Points
10 years ago

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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