How do you condense #(1/3) (2log_B M - log _B N - log_B P)#?

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How do you condense #(1/3) (2log_B M - log _B N - log_B P)#?

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Answer 1 · 2016-05-22T23:14:19

It is

#(1/3) (2log_B M - log _B N - log_B P)= (1/3)*(logM^2/logB+logN^-1/logB+logP^-1/logB)= (1/3)((logM^2+logN^-1+logP^-1)/logB)= (1/3)(log(M^2/(N*P))/logB)= (log(root 3 (M^2/(N*P)))/logB)#

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How do you condense #(1/3) (2log_B M - log _B N - log_B P)#?

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