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Silvania012
@Silvania012
December 2019
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Sabendo que log2=0,30, log3=0,48 e log 7=0,85, calcule:
a) Log 56
b) Log 49
c) Log12
d) Log 30
e) Log 0,7
f) Log 400
g) Log² 9
h) Log7 100
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andrefernandes2
Tire o mmc de 56
56------2
28------2
14-------2
7--------7
1
log2 elevado a 3 - log7
3 log2 + log7
3×0,30 + 0,85
0,9+0,85= 1,75
b)49-----7
7------7
1
log7 elevado a 2
2 log 7
2×0,85=1,7
c) 12------2
6-----2
3----3
1
2log de 2 + log3
2×0,30 + 0,48
0,6+0,48=1,08
d) 30-------10
3-------3
1
1 log10 + 1 log3
1+0,48 = 1,48
e) log 7/ log 10
log7- log10
0,85-1
-0,15
f) 400----10
40----10
4-----2
2-----2
1
2 log10+ 2 log 2
2×1+2×0,30
2+0,60 = 2,60
g) 9----3
3----3
1
log de 2 elevado a 3
3 log de 2
3× 0,30
0,90
h) 100---10
10-----10
1
log7elevado a 2
2 log de 7
2 ×0,85= 1,7
1 votes
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silvania012
Obrigada por ter me ajudado!
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56------2
28------2
14-------2
7--------7
1
log2 elevado a 3 - log7
3 log2 + log7
3×0,30 + 0,85
0,9+0,85= 1,75
b)49-----7
7------7
1
log7 elevado a 2
2 log 7
2×0,85=1,7
c) 12------2
6-----2
3----3
1
2log de 2 + log3
2×0,30 + 0,48
0,6+0,48=1,08
d) 30-------10
3-------3
1
1 log10 + 1 log3
1+0,48 = 1,48
e) log 7/ log 10
log7- log10
0,85-1
-0,15
f) 400----10
40----10
4-----2
2-----2
1
2 log10+ 2 log 2
2×1+2×0,30
2+0,60 = 2,60
g) 9----3
3----3
1
log de 2 elevado a 3
3 log de 2
3× 0,30
0,90
h) 100---10
10-----10
1
log7elevado a 2
2 log de 7
2 ×0,85= 1,7