Question

log3 (x - 1) + 2 log9 (17+x) = 7 + log1/3 9

Answer 1

$log_3 (x - 1) + 2 log_9 (17+x) = 7 + log_{1/3} 9$

$log_3 (x - 1) + log_3 (17+x) = 7 - log_{3} 3^2$

$log_3 [(x - 1) *(17+x)] = 5$

$3^5 = (x - 1) *(17+x)$

$-243+ left( x-1 right) left( 17+x right) =0$

$-243+x 17+x^2-17-x=0$

$x^2+16 x-260 = 0$

$x_{1,2}= frac{-bpmsqrt{D}}{2a} = frac{-16pmsqrt{1296}}{2} = frac{-16pm36}{2}$

$x_1 = 10,; x_2 = -26$

ОДЗ:
$left { {{x - 1>0} atop {17+x>0}} right.$
$left { {{x >1} atop {x>-17}} right.$

Ответе:$x = 10$