Question

lg(х+1) + lg(х-1) = lg3

Answer 1

одз:
x+1>0
x-1>0
x>-1
x>1
$x in (1;+infty)$
решаем:
$lg(x+1) +lg(x-1) = lg3 \lg(x+1)(x-1)=lg3 \lg(x^2-1)=lg3 \x^2-1=3 \x^2=4 \x_1=2 in (1;+infty) \x_2=-2 notin (1;+infty)$
Ответ: x=2