Question

Решить неравенство:

 

log5(x-3)+log5(x+1)=1

Answer 1

$\log_5(x-3)+log_5(x+1)=1\ x-3>0 wedge x+1>0\ x>3 wedge x>-1\x>3\ log_5(x-3)(x+1)=1\ log_5(x^2+x-3x-3)=1\ log_5(x^2-2x-3)=1\ 5^1=x^2-2x-3\ x^2-2x-8=0\ x^2+2x-4x-8=0\ x(x+2)-4(x+2)=0\ (x-4)(x+2)=0\ x=4 vee x=-2 \\ underline{x=4}$