Question

Dados log2= 0,3 ; log5= 0,7 ; log11= 1,04 e log3= 0,48 , Calcule:
a) log110
b) log60
c) log99
d) log0,32

Answer 1

Resposta:

• Dados:

$Log2=0,3$

$Log5=0,7$

$Log11=1,04$

$Log3=0,48$

• Sabemos que:

$Log_{10}x=y$

$Log(xy)=Log(x)+Log(y)$

$Log(\frac{x}{y} )=Log(x)-Log(y)$

$Log_{x}x=1$

$Log_ab=\frac{Log_cb}{Log_ca}$

Explicação passo-a-passo:

A)

$Log110$

$Log11*10$

$Log11+Log10$

$1,04+1$

$2,04$

B)

$Log60$

$Log6*10$

$Log6+Log10$

$Log(3*2)+1$

$Log3+Log2+1$

$0,48+0,3+1$

$1,78$

C)

$Log99$

$Log33*3$

$Log(11*3)+Log3$

$Log11+Log3+Log3$

$1,04+0,48+0,48$

$2$

D)

$Log0,32$

$Log\frac{32}{100}$

$Log32-Log100$

$Log2^5-Log10^2$

$Log2*2*2*2*2+Log10*10$

$Log2+Log2+Log2+Log2+Log2+Log10+Log10$

$0,3+0,3+0,3+0,3+0,3+1+1$

$3,5$

Espero ter ajudado!!

Answer 2

\colorbox{blue}{\color{white}{}}

a)

$\colorbox{blue}{\color{white}{ log(110) = log(11.10)}} \\ \colorbox{blue}{\color{white}{ log(110) = log(11) + log(10)}}$

$\colorbox{blue}{\color{white}{ log(110) =1,04 +1 =2,04 }}$

b)

$\colorbox{blue}{\color{white}{ log(60) = log(2.3.10)}} \\ \colorbox{blue}{\color{white}{= log(2) + log(3) + log(10)}} \\ \colorbox{blue}{\color{white}{= 0,3 + 0,48 + 1 = 1,78}}$

c)

$\colorbox{blue}{\color{white}{ log(99) = log(3.3.11)}} \\ \colorbox{blue}{\color{white}{ = log(3) + log(3) + log(11)}}$

$\colorbox{blue}{\color{white}{ log(99) =0,3 + 0,3 + 1,04 }} \\ \colorbox{blue}{\color{white}{ log(99) =1,64}}$

d)

$\color{orangered}{ log(0,32)} \\ \color{orangered}{ = log( \frac{8}{25}) = log(8) - log(25)} \\ = \color{orangered}{ log( {2})^{3}} - log( {5})^{2} \\ = \color{orangered}{3 log(2) - 2 log(5) }$

$\colorbox{blue}{\color{white}{ log(0,32) } = 3.0,3 - 2.0,7 } \\\colorbox{blue}{\color{white}{ log(0,32) } = 0,9 -1,4 = - 0,5 }$