Question

daca se poate rezolvati 70a si 71 va rog!!!
dau coroana!!

Answer 1

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

$t_{1} =\frac{4+2\sqrt{5} }{2} =2+\sqrt{5}=2^{x} \\ t_{2}=2-\sqrt{5} =2^{x}$

lg(2+√5)=lg2ˣ ⇒$x_{1}=\frac{lg(2+\sqrt{5} )}{lg2} \\x_{2}=\frac{lg(2-\sqrt{5} )}{lg2}$