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If (log)k xdot(log)5k=(log)x5,k!=1,k >0,...

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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`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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CENGAGE ENGLISH-LOGARITHM-All Questions
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