Răspuns:
Explicație pas cu pas:
70. a. f(x)=(log₂(3x))²-4(log₂(6x))+7
f(x)=(log₂(3x))²-4(log₂(2×3x))+7
f(x)=(log₂(3x))²-4(log₂2+log₂(3x))+7
Notam log₂(3x)=y
f(x)=y²-4(1+y)+7
f(x)=y²-4y+3=g(y) functie de gradul II
71. a=lg2
b=lg(2ˣ-1)
c=lg(2ˣ+1)
2b=a+c (3 termeni consecutivi ai unei progresii aritmetice )
2lg(2ˣ-1)=lg2+lg(2ˣ+1)
lg(2ˣ-1)²=lg[2(2ˣ+1)] scapam de logaritm
Notam 2ˣ=t
(t-1)²=2t+2
t²-2t+1-2t-2=0
t²-4t-1=0
Δ=16+4=20
[tex]t_{1} =\frac{4+2\sqrt{5} }{2} =2+\sqrt{5}=2^{x} \\ t_{2}=2-\sqrt{5} =2^{x}[/tex]
lg(2+√5)=lg2ˣ ⇒[tex]x_{1}=\frac{lg(2+\sqrt{5} )}{lg2} \\x_{2}=\frac{lg(2-\sqrt{5} )}{lg2}[/tex]
