How do you calculate ${log}_{8} 512$ $log_8 512$ ?

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Answer 1 · 2016-04-16T20:14:23

Using logarithm laws we have that

$log_8 512=log512/(log8)=log2^9/(log2^3)=(9*log2)/(3*log2)= (9*cancellog2)/(3*cancellog2)=9/3=3$

Finally

$log_8 512=3$

How do you calculate log_8 512log8512log_8 512?