Решите уравнение (log(3)√x+8) – 1 =log(3)2-log(3)√x-8

(log₃√(x+8)) – 1 = log₃2-log₃√(x-8)   одз   х+8> 0   x > - 8   x-8> 0 x > 8

1/2log₃(x+8) – 1 = log₃2 - 1/2log₃ (x-8)

1/2log₃(x+8) +1/2log₃ (x-8) = log₃2 +1

1/2 *( log₃(x+8) +log₃ (x-8) = log₃2 +log₃3

log₃(x+8) *(x-8) = 2* log₃2*3

log₃(x²-8² ) = log₃6²

х²-64=36

х²=100

х₁=10   х₂= -10

ответ с учетом одз х=10