Respuesta :

[tex]2(\log 18-\log 3)+\dfrac 12 \log \dfrac{1}{16}=2 \log \dfrac{18}{3}+\log \sqrt{\dfrac{1}{16}}\\=2 \log 6+\log \dfrac{1}{4}=\log 6^2+\log \dfrac{1}{4}=\log 36+\log \dfrac{1}{4}\\=\log \dfrac{36}{4}=\log 9[/tex]

beinggreat78

Answer:

1.556 or [tex]\mathrm{2\log _{10}\left(6\right)}[/tex]

Step-by-step explanation:

[tex]2\left(\log _{10}\left(18\right)-\log _{10}\left(3\right)\right)+\frac{1}{2}\cdot \frac{\log _{10}\left(1\right)}{16}\\\log _{10}\left(18\right)-\log _{10}\left(3\right)\\ = 18 \div 3\\= 6\\= log_{10}(6)\\\\\log _{10}\left(18\right)-\log _{10}\left(3\right) = log_{10}(6)\\\\\frac{1}{2} \cdot\frac{\log _{10}\left(1\right)}{16}\\log_{10}(1) = 0\\= \frac{1}{2} \cdot\frac{0}{16}\\\\0 \div 16 = 0\\a \cdot 0 = 0\\= 0\\\\= 2\log _{10}\left(6\right)+0\\=2\log _{10}\left(6\right)[/tex]

Hope this helps!

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