How do you condense #1/3log 3x + 2/3log 3x #?

1 Answer
Bdub
Mar 22, 2016

#1/3 log 3x + 2/3 log 3x##=log(3x)#

Explanation:

#1/3 log 3x + 2/3 log 3x#

#= log(3x)^(1/3) + log(3x)^(2/3)#-> use property #log_bx ^n = nlog _b x#

#=log((3x)^(1/3) (3x)^(2/3))#-> use property #log_b(xy)=log_bx+log_by#

#=log((3x)^(1/3+2/3))#-> use property #x^a*x^b=x^(a+b)#

#=log(3x)#

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