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Chacha33
@Chacha33
May 2019
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Bonjour pouvez vous m'aidez à résolvez ces équations s'il vous plait, merci beaucoup!
eײ⁻³ = 2
et
eײ⁻³ = -2
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no63
Salut
e^x²-3=2
=> ln(e^x²-3)=ln(2)
=> x²-3=ln2
=> x²=3+ln2
=> x= racine(3+ln2) ou x= - racine(3+ln2)
e^x²-3= -2 pas possible car e^x>0
1 votes
Thanks 1
Commentaires (3)
1) exp(x^(2-3))=2
exp(x^-1)=2
exp(1/x)=2
1/x=ln(2)
x=1/ln(2)
2) exp(x^(2-3))=-2
impossible
! car exp(y)>0
0 votes
Thanks 0
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e^x²-3=2
=> ln(e^x²-3)=ln(2)
=> x²-3=ln2
=> x²=3+ln2
=> x= racine(3+ln2) ou x= - racine(3+ln2)
e^x²-3= -2 pas possible car e^x>0
exp(x^-1)=2
exp(1/x)=2
1/x=ln(2)
x=1/ln(2)
2) exp(x^(2-3))=-2
impossible ! car exp(y)>0