Question

решите уравнение log3 (3^x-8)=2-x

Answer 1

$\log_3 (3^x-8)=2-x\ 3^x-8>0\ 3^x>8\ x>log_38\ 3^{2-x}=3^x-8|cdot3^x\ 3^2=(3^x)^2-8cdot3^x\ (3^x)^2-8cdot3^x-9=0\ (3^x)^2+3^x-9cdot3^x-9=0\ 3^x(3^x+1)-9(3^x+1)=0\ (3^x-9)(3^x+1)=0\ 3^x-9=0\ 3^x=9\ x=2$