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GabrielCS19
@GabrielCS19
January 2020
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Resolva a inequação 2 elevado a 3x-5 > 128 .
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emicosonia
Resolva a inequação 2 elevado a 3x-5 > 128 .
deixar as bases iguais
2³ˣ ⁻ ⁵ > 128 ( 128 = 2x2x2x2x2x2x2 = 2⁷)
2³ˣ⁻⁵ = 2⁷ BASES IGUAIS
3x - 5 = 7
3x = 7 + 5
3x = 12
x = 12/3
x = 4 ( resposta)
também podemo fazer
128| 2
64| 2
32| 2
16| 2
8| 2
4|2
2/2
1/ = 2.2.2.2.2.2.2 = 2⁷
3 votes
Thanks 3
GabrielCS19
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Lista de comentários
deixar as bases iguais
2³ˣ ⁻ ⁵ > 128 ( 128 = 2x2x2x2x2x2x2 = 2⁷)
2³ˣ⁻⁵ = 2⁷ BASES IGUAIS
3x - 5 = 7
3x = 7 + 5
3x = 12
x = 12/3
x = 4 ( resposta)
também podemo fazer
128| 2
64| 2
32| 2
16| 2
8| 2
4|2
2/2
1/ = 2.2.2.2.2.2.2 = 2⁷