If log2= a and log3= b evaluate log(0.375)?

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Answer 1 · 2017-03-24T09:20:53

$log0.375=b-3a$

Note that $log5=log(10/2)=log10-log2=(1-a)$

We will now work on $log0.375$ .

$log0.375$

= $log(375/1000)$

= $log375-log1000$

= $log(3xx5xx5xx5)-3$

= $log3+3log5-3$

= $b+3(1-a)-3$

= $b+3-3a-3$

= $b-3a$

Alternatively , one can also write

$log0.375=log(3/8)=log3-log8=log3-log2^3$

= $log3-3log2=b-3a$