How do you prove #-4log3 + log 3 = log(1/27)#?

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How do you prove #-4log3 + log 3 = log(1/27)#?

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Answer 1 · 2016-03-22T19:25:12

#-4log3+log3=log(1/27)#

Explanation:

#-4log3+log3=log(1/27)#
#LHS:-4 log3+log3#
#=-3log3#
#=log 3 ^-3#-> use property #log_bx^n=n*log_bx#
#=log(1/3^3)#-> use property #x^-n=1/x^n#
#=log (1/27)#
#=RHS#

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How do you prove #-4log3 + log 3 = log(1/27)#?

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