Fala ae Murilo,
boa noite!
De acordo com a tabela: [tex]\begin{cases} \log 2 = 0,30 \\ \log 3 = 0,47 \end{cases}[/tex]
Segue,
[tex]\\ \log 7,2 + \log \sqrt{3} = \\ \log \left ( 7,2 \cdot \sqrt{3} \right ) = \\\\ \log \left (\frac{72\sqrt{3}}{10} \right ) = \\ \log 72\sqrt{3} - \log 10 = \\\\ \log \left ( 2^3 \cdot 3^2 \cdot 3^{\frac{1}{2}} \right ) - \log 10^1 = \\\\ \log \left ( 2^3 \cdot 3^{\frac{5}{2}} \right ) - 1 = \\\\ \log 2^3 + \log 3^{\frac{5}{2}} - 1 = \\\\ 3 \cdot \log 2 + \frac{5}{2} \cdot \log 3 - 1 = \\\\ 3 \cdot 0,30 + \frac{5}{2} \cdot 0,47 - 1 = \\ \boxed{\boxed{1,075}}[/tex]
