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Biancaas
@Biancaas
April 2020
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O valor de log 128 na base 2 + log 8 na base 4 é?
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DANILOCICILIOTTI
Log2 128 + log4 8
por partes
2^x = 128 colocando na mesma base
2^x = 2^7
x=7
4^x = 8
2^2+x = 2^3
resolvendo os expoentes
2+x= 3
x= 1
Então 7+1 = 8
1 votes
Thanks 1
ittalo25
(^ = elevado)
log 128 na base 2 + log 8 na base 4
128 = 2^x
2^7 = 2^x
x = 7
log 8 na base 4
8 = 4^x
2^3 = 2^2x
3 = 2x
x = 3/2
7 + 3/2 =
17/2
2 votes
Thanks 1
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Lista de comentários
por partes
2^x = 128 colocando na mesma base
2^x = 2^7
x=7
4^x = 8
2^2+x = 2^3
resolvendo os expoentes
2+x= 3
x= 1
Então 7+1 = 8
log 128 na base 2 + log 8 na base 4
128 = 2^x
2^7 = 2^x
x = 7
log 8 na base 4
8 = 4^x
2^3 = 2^2x
3 = 2x
x = 3/2
7 + 3/2 = 17/2