How can we express log12 in terms of log2 and log3?

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Answer 1 · 2018-05-03T15:19:41

#log12 = 2log2 + log3#

Assuming you mean:

#log12 = log_a 12#

where #a# is some base.

Then we can use product rule for logarithms

#=>log_a(x*y) = log_a x + log_a y#

Hence,

#log12 = log(4*3) = log(2*2*3)#

#log12= log2 + log2 + log3#

#log12 = 2log2 + log3#