Considerando log2= 0,3010 e log3= 0,4771, calcule
a) log 72
b) log 200
c) log raiz de 108
d) log 5 raiz de 5
Respostas
respondido por:
3
log 2 = 0,3010
log 3 = 0,4771
log 5 = 0,6989
a)log 72 = log 2³.3² = log2³ + log3² =3 log 2 + 2 log 3 = 3(0,3010 ) + 2(0,4771) = 0,903 + 0,9542 = 1,8572
72|2
36|2
18|2
9 |3
3 |3
1
b)log 200 = log 2³. 5² = 3 log 2 + 2 log 5 = 3 (0,3010) + 2 (0,6989) = 0,903 + 1,3978 = 2,3008
200|2
100|2
50|2
25|5
5 |5
1 ]
c) log √108 = log √2².3³ =log √4 . √3².√ 3 = log 2 . 3. √3= log 6√3 =
= log 6 . 3^1/2 = log 2 . 3 . 3^1/2 = log 2 + log 3 + 1/2 log 3 =
=0,3010 + 0,4771 + 1/2(0,4771 = 0,3010/1 + 0,4771/1 +0,4771/2 =
= 0,602/2 + 0,9542/2 + 0,4771/2 = 2,0333/2 = 1,01665
d) log 5 √5 = log 5 + log √5 = log 5 + log 5 ^1/2 = log 5 + 1/2 log 5 =
=0,6989 + 1/2 (0,6989) = 0,6989 / 1 + 0,6989/2 = 1,3978/2 + 0,6989/2 =
2,0967/2 = 1,04835
log 3 = 0,4771
log 5 = 0,6989
a)log 72 = log 2³.3² = log2³ + log3² =3 log 2 + 2 log 3 = 3(0,3010 ) + 2(0,4771) = 0,903 + 0,9542 = 1,8572
72|2
36|2
18|2
9 |3
3 |3
1
b)log 200 = log 2³. 5² = 3 log 2 + 2 log 5 = 3 (0,3010) + 2 (0,6989) = 0,903 + 1,3978 = 2,3008
200|2
100|2
50|2
25|5
5 |5
1 ]
c) log √108 = log √2².3³ =log √4 . √3².√ 3 = log 2 . 3. √3= log 6√3 =
= log 6 . 3^1/2 = log 2 . 3 . 3^1/2 = log 2 + log 3 + 1/2 log 3 =
=0,3010 + 0,4771 + 1/2(0,4771 = 0,3010/1 + 0,4771/1 +0,4771/2 =
= 0,602/2 + 0,9542/2 + 0,4771/2 = 2,0333/2 = 1,01665
d) log 5 √5 = log 5 + log √5 = log 5 + log 5 ^1/2 = log 5 + 1/2 log 5 =
=0,6989 + 1/2 (0,6989) = 0,6989 / 1 + 0,6989/2 = 1,3978/2 + 0,6989/2 =
2,0967/2 = 1,04835
everllycarolinp11yz5:
muito obgd ... me ajudou bastante
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