Răspuns :
Răspuns:
conditii de existenta:
3 numar pozitiv,≠1
x²+1 >0
log3 (x²+1)=log 3 (5)
x²+1=5
x²=√4
X=±2
[tex]\displaystyle\bf\\log_3(x^2+1)=log_3\,5\\\\Logaritmii~au~aceeasi~baza.\\\\\implies~x^2+1=5\\\\x^2=5-1\\x^2=4\\x_{12}=\pm\sqrt{4}\\\\ \boxed{\bf x_1=-2~~si~~x_2=2}[/tex]