Question

Rez ec : log[x-1]x+2=2

Answer 1

am notat cu " ' " : din baza
log'(x-1) x+2 = 2
C.E : (1) x-1 >0
x € ( 1,+inf)

x-1 diferit 0
x diferit 1

(2)x+2>0
x € (-2,+inf)
D: x € (1,+inf)

log'(x-1) x+2 = 2 (inj)
x+2=(x-1)^2
x+2=x^2-2x+1
x^2-3x-1=0
d=9-4=5 rad(d)=rad(5)
x12=(3+/-rad(5) )/2 ,x2 € D
S: x € {(3+rad(5) )/2}

Answer 2

[tex]log_\Big{x-1}x+2 = 2 \\ \\ $Conditii de existenta: $ \\ \left\{ \begin{array}{c} x-1 \ \textgreater \ 0 \\ x-1 \neq 1\\ x+2 \ \textgreater \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{c} x \ \textgreater \ 1 \\ x \neq 2\\ x \ \textgreater \ -2 \end{array} \right |\Rightarrow D = (1,2)\cup(2,+\infty) \\ \\ \\ log_\Big{x-1}x+2 = 2 \Rightarrow x+2 = (x-1)^2 \Rightarrow x+2 = x^2-2x+1 \Rightarrow \\ \Rightarrow x^2-2x-x+1-2 = 0 \Rightarrow x^2-3x-1=0 [/tex]
[tex]\Delta = (-3)^2-4\cdot 1\cdot 1 = 9-4 = 5 \Rightarrow x_{1,2} = \dfrac{-(-3)\pm \sqrt{5} }{2\cdot 1} \Rightarrow \\ \Rightarrow x_{1,2} = \dfrac{3 \pm \sqrt{5} }{2} \Rightarrow \left| \begin{array}{c} x_1 = \dfrac{3- \sqrt{5} }{2} \ \textless \ 1 \notin D \\ x_2 = \dfrac{3+ \sqrt{5} }{2} \in D \end{array} \right |\Rightarrow \boxed{S = \left\{\dfrac{3+ \sqrt{5} }{2} \right\}}[/tex]