Question

Sa se rezolve ecuațiile (logaritmice). Urgent vă rog! Dau coroană! ​

Answer 1

 

$\displaystyle\\e)\\ \log_{x+1}5=1\\\\(x+1)^1=5\\\\x+1=5\\\\x=5-1\\\\\boxed{x=4}\\\\\\f)\\\log_{x+2}\left(2x^3+5x+2\right)=1\\\\(x+2)^1=2x^3+5x+2\\\\2x^3+5x+2=x+2\\\\2x^3+5x+2-x-2=0\\\\2x^3+4x=0~~~\Big|:2\\\\x^3+2x=0\\\\x(x^2+2)=0\\\\x=0~sau~x^2+2 = 0\\\\\boxed{x1=0}\\\\x^2=-2\\\\x_{23}=\pm\sqrt{-2}\\\\ \boxed{x_2=i\sqrt{2}}\\\\ \boxed{x_3=-i\sqrt{2}}$

$\displaystyle\\g)\\\log_{x-1}\left(2x^2-7x+1\right)=2\\\\(x-1)^2=2x^2-7x+1\\\\2x^2-7x+1=(x-1)^2\\\\2x^2-7x+1=x^2-2x+1\\\\2x^2-7x+1-x^2+2x-1=0\\\\x^2-5x=0\\\\x(x-5)=0\\\\x=0~sau~x-5=0\\\boxed{x_1=0}\\\\\boxed{x_2=5}$

Answer 2

f)

$\it log_{x+2}(2x^3+5x+2) =1 \Rightarrow 2x^3+5x+2 =x+2 \Rightarrow 2x^3+4x=0 \Rightarrow\\ \\ \Rightarrow x(x^2+2) =0\ \ \ (1)\\ \\ x^2\geq0|_{+2} \Rightarrow x^2+2\geq 2 > 0\ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow x=0$

După verificare (etapă necesară dacă nu am precizat domeniul de existență),

avem:

$\it x=0 \Rightarrow log_2 2=1 \Rightarrow1=1\ (A)\\ \\ Prin\ urmare,\ ecua\c{\it t}ia\ dat\breve{a}\ admite\ solu\c{\it t}ia\ unic\breve{a} \ \ x=0.$

e)

$\it log_{x-1} (2x^2-7x+1)=2$

O primă condiție de existență a ecuației este :

$\it x-1>0 \Rightarrow x>1\ \ \ \ (*)$

Ecuația dată se poate transforma astfel:

$\it 2x^2-7x+1=(x-1)^2 \Rightarrow 2x^2-7x+1=x^2-2x+1 \Rightarrow x^2-5x=0\Rightarrow\\ \\ \Rightarrow x(x-5) =0 \stackrel{(*)}{\Longrightarrow} x=5$

Verificare:

$\it x=5 \Rightarrow log_4(50-35+1)=2 \Rightarrow log_4 16 =2 \Rightarrow 2=2\ (A)$

Așadar, ecuația dată admite soluția unică  $\it x=5$