The value of #x# satisfying the equation?

#1/log_3(1994)+1/log_6(1994)+1/log_9(1994)+1/log_12(1994)=1/(log_(3x)(1994))#

1 Answer
P dilip_k
Jul 16, 2018

#1/log_3(1994)+1/log_6(1994)+1/log_9(1994)+1/log_12(1994)=1/(log_(3x)(1994))#

#=>log_(1994)3+log_(1994)6+log_(1994)9+log_(1994)12=log_(1994)(3x)#

#=>log_(1994)(3xx6xx9xx12)=log_(1994)(3x)#

#=>3x=3xx6xx9xx12#

#=>x=648#

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