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

Viranch Mistry
32 Points
```										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 !! :)
```
7 years ago
Anish Nair
19 Points
```										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
```
7 years ago
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