The value of $x$ satisfying the equation?

Question

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

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Answer 1 · 2018-07-16T14:03:26

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

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

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

$=>3x=3xx6xx9xx12$

$=>x=648$