Resposta:
1. b
2.e
3. e
4. d
5. 3
Explicação passo a passo:
1.
[tex]log (28) = log (2\cdot 2\cdot7)=log(2)+log(2)+log(7)\\log(28) = 0.301 + 0.301+0.845 = 1.447[/tex]
2.
[tex]log (\sqrt[10]{3200} ) = log(3200^{\frac{1}{10} }) = \dfrac{1}{10} \cdot log(3200)\\log (\sqrt[10]{3200} ) = \dfrac{log(5^2\cdot2^7)}{10} = \dfrac{2\cdot log(5)+7\cdot log(2)}{10}[/tex]
[tex]log(5)= log (10) - log(2) = 1-0.301 = 0.699[/tex]
[tex]log (\sqrt[10]{3200} ) = \dfrac{2\cdot0.699+7\cdot0.301}{10} = 0.3505[/tex]
3.
[tex]\text {PH} = log \left(\dfrac{1}{H^+}\right) \\\text {PH} = log \left(\dfrac{1}{2.5\cdot10^{-9}}\right) = log \left(\dfrac{1}{25\cdot10^{-10}}\right)\\[/tex]
[tex]\text {PH} = log(1) - (log(25)-10\cdot log(10))[/tex]
[tex]\text {PH} = 0 - (2\cdot log(5) - 10) = -(1.4-10) = 8.6[/tex]
4.
O ponto (0,1) não faz parte da função [tex]f(x) = log_2(x)[/tex] porque [tex]2^1\neq 0[/tex]
5.
[tex]log(25)+log(5)+log(4)+log(2)\\2\cdot log(5) + log(5) + 2\cdot log(2) + log(2)\\3\cdot log(5) + 3\cdot log (2)\\3(1-log(2))+3\cdot log(2)\\3 - 3\cdot log(2) + 3 \cdot log(2) = 3[/tex]
