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If `(log)_k xdot(log)_5k=(log)_x5,k!=1,k >0,`then `x`is equal to`k`(b) 1/5(c) 5 (d) none of these |
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Answer» `log_y^x=log_c^x/log_c^y` `(logx/logk)*(logk/log5)=log5/logx` `(logx)^2=(log5)^2` `logx=log5` `x=5`. |
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