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

1 Answer
Mar 24, 2017

Answer:

#log0.375=b-3a#

Explanation:

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#