Log3 (x - 1) + 2 log9 (17+x) = 7 + log1/3 9

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

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

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

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

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

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

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

$x_{1,2}= \frac{-b\pm\sqrt{D}}{2a} = \frac{-16\pm\sqrt{1296}}{2} = \frac{-16\pm36}{2}$

$x_1 = 10,\; x_2 = -26$

ОДЗ:

0} \atop {17+x>0}} \right. " alt=" \left \{ {{x - 1>0} \atop {17+x>0}} \right. " align="absmiddle" class="latex-formula">

1} \atop {x>-17}} \right. " alt=" \left \{ {{x >1} \atop {x>-17}} \right. " align="absmiddle" class="latex-formula">

Ответе: $x = 10$