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

log_zxy= c-1   =>  log_zxy +1= c  =>  log_zxy + log_zz = c   =>  log_zxyz = c

Thus we have, c = log_zxyz,  a= log_xxyz  and   b= log_yxyz

Now, LHS =   ab + bc +ca  

let us first consider 'ab' ..... ab =  (log_xxyz).(log_yxyz)  =  [(log_exyz)/(log_ex)] . [(log_exyz)/(log_ey)]

                                                                            =  (log_exyz)² / logx.logy

Simlarly find bc and ac and add you will get     (log_exyz)³ / logx.logy.logz

And RHS = abc = (log_exyz)³ / logx.logy.logz

Therefore, LHS = RHS !! :)

c-1=Log_zxy

c=Log_zxy +1

c= Log_zxy+Log_zz

c= Log_z(xyz) --1

Similarly,

b= Log_yxyz

a= Log_xxyz 

put the values of a b c in given equation and get the result