Resposta:

Os cálculos usando as propriedades operatórias dos logaritmos são mostrados abaixo.

Explicação passo a passo:

[tex]a)\;\begin{array}{ccl}\log 50&=&\log\; (2\;.\;25)\\&=&\log 2 + \log 25\\&=&\log 2 + \log 5^2\\&=&\log 2 + 2\;.\;\log 5\\&=&\log 2 + 2\;.\;\log \frac{10}{2}\\&=&\log 2 + 2\;.\;(\log 10 - \log 2)\\&=&0{,}301+2\;.\;(1 - 0{,}301)\\&=&0{,}301+2\;.\;0{,}699\\&=&0{,}301+1{,}398\\&=&1{,}699\end{array}[/tex]

[tex]b)\;\log 512=\log 2^9=9\;.\;\log 2=9\;.\;0{,}301=2{,}709\\\\\\[/tex]

[tex]c)\;\begin{array}{ccl}\log 720&=&\log\; (2^4\;.\;3^2\;.\;5)\\&=&4\;.\;\log 2 + 2\;.\;\log 3+\log 5\\&=&4\;.\;\log 2 + 2\;.\;\log 3+\log \frac{10}{2}\\&=&4\;.\;\log 2 + 2\;.\;\log 3+ \log 10 - \log 2\\&=&4\;.\;0{,}301+2\;.\;0{,}477+1-0{,}301\\&=&1{,}204+0{,}954+1-0{,}301\\&=&2{,}857\end{array}[/tex]

[tex]d)\;\log \sqrt{27}=\log \sqrt{3^3}=\log 3^{\frac{3}{2}}=\frac{3}{2}\;.\;\log 3=\frac{3}{2}\;.\;0{,}477=0{,}7155[/tex]

[tex]e)\;\begin{array}{ccl}\log \frac{1}{18}&=&\log 18^{-1}\\&=&-1\;.\;\log 18\\&=&-1\;.\;\log (2\;.\;3^2)\\&=&-1\;.\;(\log 2 + 2\;.\;\log 3)\\&=&-1\;.\;(0{,}301+2\;.\;0{,}477)\\&=&-1\;.\;(0{,}301+0{,}954)\\&=&-1\;.\;1{,}255\\&=&-1{,}255\end{array}[/tex]

[tex]f)\;\begin{array}{ccl}\log\;(4\;.\;\sqrt{6})&=&\log 4 +\log \sqrt{6}\\\\&=&\log 2^2+\log 6^{\frac{1}{2}}\\\\&=&2\;.\;\log 2+\frac{1}{2}\;.\;\log 6\\\\&=&2\;.\;\log 2+\frac{1}{2}\;.\;\log\;(2\;.\;3)\\\\&=&2\;.\;\log 2+\frac{1}{2}\;.\;(\log\;2+\log 3)\\\\&=&2\;.\;0{,}301+\frac{1}{2}\;.\;(0{,}301+0{,}477)\\\\&=&0{,}602+\frac{1}{2}\;.\;0{,}778\\\\&=&0{,}602+0{,}389\\\\&=&0{,}991\end{array}[/tex]

[tex]g)\;\begin{array}{ccl}\log\;(9\;.\;\sqrt{2})&=&\log 9 +\log \sqrt{2}\\\\&=&\log 3^2+\log 2^{\frac{1}{2}}\\\\&=&2\;.\;\log 3+\frac{1}{2}\;.\;\log 2\\\\&=&2\;.\;0{,}477+\frac{1}{2}\;.\;0{,}301\\\\&=&0{,}954+0{,}1505\\\\&=&1{,}1045\end{array}[/tex]