Question

Buna ziua ,
Am si eu nevoie de ajutor la:
Aflati solutia reala a ecuatiei $log 3 \frac{x-5}{x+3}=2$.
Multumesc,

Answer 1

[tex]9= \frac{x-5}{x+3} 9(x+3)=x-5 9x+27=x-5 9x+27-x+5=0 8x+32=0 8x=-32 x=-4 [/tex]Conform formulei $log_{a}^{b}=c =)))) a^c=b$

[tex] 3^{2}= \frac{x-5}{x+3} [/tex]

Answer 2

$\displaystyle log_3 \frac{x-5}{x+3} =2 \\ log_3 \frac{x-5}{x+3}=log_33^2 \\ \frac{x-5}{x+3} =3^2 \\ \frac{x-5}{x+3} =9 \\ x-5=9(x+3) \\ x-5=9x+27 \\ x-9x=27+5 \\ -8x=32 \\ x=- \frac{32}{8} \\ x=-4$