9514 1404 393
Answer:
log₁₂(4)
Step-by-step explanation:
The applicable rule of logarithms (to the same base) is ...
log(a^b) = b·log(a)
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Then the expression simplifies as follows:
[tex]3\log_{12}(2)+\dfrac{1}{3}\log_{12}(2^3)-\log_{12}(2^2)\\\\=3\log_{12}(2)+\dfrac{3}{3}\log_{12}(2)-2\log_{12}(2)=(3+1-2)\log_{12}(2)\\\\=2\log_{12}(2)=\log_{12}(2^2)=\boxed{\log_{12}(4)}[/tex]
