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If (log x)/(2a+3b-5c)=(log y)/(2b+3c-5a)=(log z)/(2c+3a-5b) then xyz= a.2 b.1 c.0 d.-1
Given:- logx/2a+3b-5c=logy/2b+3c-5a=logx/2c+3a-5bTherefore, we can write it as logx+logy+ logz/0So, logx+ logy+ logz=0So,log(xyz)=0xyz=1
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