Question

13) Se log2=0,30; log3=0,48 e log7=0,84. Determine o valor de
a) log5
f) o PH de uma solução aquosa de ácido clorídrico 0,4 molar e grau de ionização 75%
obs: restante das perguntas na foto
14) Use os dados do exercício 13 para resolver as equeções exponenciais
a) 4^x=7
b)5^x=21
c)3^x=20
d)7^x=150

Answer 1

a)$\log _{ }\left(5\right)=\log _{ }\left(\frac{10}{2}\right)=\log _{ }\left(10\right)-\log _{ }\left(2\right)=1-0,30=0,70$

b)$\log _{15}\left(21\right)=\frac{\log \left(21\right)}{\log \left(15\right)}=\frac{\log \left(7\cdot 3\right)}{\log \left(\frac{30}{2}\right)}=\frac{\log \left(7\right)+\log \left(3\right)}{\log \left(30\right)-\log \left(2\right)}$
$=\frac{\log \:\left(7\right)+\log \:\left(3\right)}{\log \left(3\cdot 10\right)-\log \left(2\right)}=\frac{\log \:\:\left(7\right)+\log \:\:\left(3\right)}{\log \:\left(3\right)+\log \left(10\right)-\log \:\left(2\right)}=\frac{0,84+0,48}{0,48+1-0,30}=\frac{1,32}{1,18}$

c)$\log _{45}\left(24\right)=\frac{\log \left(24\right)}{\log \left(45\right)}=\frac{\log \left(2\cdot 2\cdot 2\cdot 3\right)}{\log \left(5\cdot 2\cdot 2\cdot 2\right)}=\frac{0.30+0,30+0,30+0,48}{0,70+0.30+0,30+0,30}=\frac{1,38}{1,6}$

d)$\log _{42}\left(350\right)=\frac{\log \left(350\right)}{\log \left(42\right)}=\frac{\log \:\left(5\cdot 7\cdot 10\right)}{\log \:\left(7\cdot 3\cdot 2\right)}=\frac{0,70+0,84+1}{0,84+0,48+0,30}=\frac{2,54}{1.62}$

e)$\log _{50}\left(0,042\right)=\log _{50}\left(\frac{42}{1000}\right)=\frac{\log \left(\frac{42}{1000}\right)}{\log \left(50\right)}$
$=\frac{\log \:\left(42\right)-\log \left(1000\right)}{\log \left(5\cdot 10\right)}=\frac{1,62-3}{0,70+1}=\frac{1,38}{1,70}$

PS: Depois edito a resposta incluindo a 14.

a)$4^x=7\rightarrow\log4^x=\log7\rightarrow\log\left(2^2\right)^x=\log7$$\rightarrow 2x\cdot\log2=\log7\rightarrow2x\cdot0,30=0,84\rightarrow$$2x=\frac{0,84}{0,30}\rightarrow x=\frac{0,84}{0,30\cdot2}\rightarrow x=\frac{0,84}{0,6}=1,4$
b)$5^x=24\rightarrow \log \left(5^x\right)=\log 24\rightarrow \:x\cdot \log 5=\log 24\rightarrow$$x\cdot \log 5=\log \:2\cdot 2\cdot 3\rightarrow \:x\cdot 0,70=0,60+0,48\rightarrow \:x=\frac{1,08}{0,70}$
c)$3^x=20\rightarrow \log 3^x=\log 20\rightarrow \:x\cdot \log 3=\log \left(10\cdot$$2\right)\rightarrow \:x\cdot 0,\:48=1+0,\:30\rightarrow \:x=\frac{1,\:30}{0,\:48}$
d)$7^x=150\rightarrow \log 7^x=\log 150\rightarrow \:x\cdot \log 7=\log 10\cdot 15\rightarrow \:x\cdot \log$$\:7=\log \left(10\right)+\log 15\rightarrow \rightarrow \:x\cdot \log 7=\log 10+\log \frac{30}{2}\rightarrow$$x\cdot \log \:7=\log 10+\log 10+\log 3-\log 2\rightarrow \:x\cdot 0,84=2+0,48-30$$\rightarrow \:x=\frac{2,18}{0,84}$