Question

1. Considerando log de 2=0,3010 e log de 3=0,4771, calcule:
a) log 5
b) log 0,0001
c) log 200
d) log 3000
e) log da raiz cúbica de 60
f) log (0,54) elevado a 0,5
2. Ache y real sabendo-se que: (imagem)

Answer 1

a)

$log 5 = log \frac{10}{2} = log10 - log2 = 1 -(0,3010) = 0699$

b) 

$log0,0001 = log10^{-4} = -4*log10 = -4*-1 = -4$

c)

$log 200 = log(2*100) = log2 + log100 = (0,3010) + 2 = 2,3010$

d)

$log 3000 = log(3*1000) = log3 + log 1000 = (0,4771) + 3 = 3,4771$

e)

$log \sqrt[3]{60} = log60^{ \frac{1}{3} } = \frac{1}{3} *log60 = \frac{1}{3} *log(2*3*10) = \\ \\ \frac{1}{3}*(log2 + log3 + log10) = \frac{1}{3} *(0,3010 +0,4771+ 1 ) = \\ \\ \frac{1,7781}{3} = 0,5927$

f)

$log (0,54)^{0,5} = log( \frac{54}{100} )^{ \frac{1}{2} } = \frac{1}{2} *(log54 - log 100) =\\ \\ \frac{1}{2} *[log(3*3*2*3) - 2] = \frac{1}{2} *(log3 + log3 + log2 + log3) - 2 = \\ \\ \frac{1}{2} *[(3*0,4771 + 0,3010 ) - 2] = \frac{1}{2} *(1,4313 + 0,3010-2) = \\ \\ \frac{1,4313+0,3010-2}{2} = \frac{1,7323-2}{2} = \frac{-0,2677}{2} = -0,13385$

2) Temos o seguinte:

$log_{2} \ y = log_{2} \ 3 + log_{2} \ 6 - 3log_{2} \ 4 \\ \\ log_{2} \ y = log_{2} \ (3*6) - log_{2} \ 4^3 \\ \\ log_{2} \ y = log_{2} \ (18) - log_{2} \ (64) \\ \\ log_{2} \ y = log_{2} \ ( \frac{18}{64} ) \\ \\ log_{2} \ y = log_{2} \ ( \frac{9}{32} ) \\ \\ y = \frac{9}{32}$