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2^2x - 3× 2^x + 1= 0=> x=?

Răspuns :

Icecon2005

[tex] 2^{2x}-3\cdot2^{x}+1=0[/tex]
aplicam:
[tex] a^{bc}=(a^{b})^{c} [/tex]
astfel avem:
[tex] 2^{2x}= (2^{x})^{2} [/tex]
[tex]\left(2^x\right)^2-3\cdot \:2^x+1=0[/tex]
notam :[tex] c [/tex]
[tex](u)^{2}-3u+1=0 [/tex]
ecuatie de gradul 2
Δ=b²-4a5=9-4=5
x₁=(-b+√Δ)/2a=(3+√5)/2
x₂=(-b-√Δ)/2a=(3-√5)/2

[tex]2^{x}=([/tex]3 + √5)/2⇒[tex]2^{x}=([/tex]3+√5)/2

[tex]2^{x}=([/tex]3 -√5)/2⇒[tex]2^{x}=([/tex]3-√5)/2

x₁=[ln(3+√5)]/ln(2)   si x₂=[ln(3+√5)]/ln(2)
Albatran
fie 2^x=t
atunci
2^(2x) =2^(x*2) =(2^x)²=t²
ecuatia devine
t²-3t+1=0
t1,2=(3+/-√(9-4))/2
t1,2=(3+/-√5)/2
cum 3>√5, t1 si t2>0
2^x=(3-√5)/2
x1= log in baza 2din ((3-√5)/2)

2^x=(3+√5)/2
x2=log in baza 2 din ((3+√5)/2)

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