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x > 0 i x różne od 1
zatem
x^2 = 6x - 9
x^2 - 6x + 9 = 0
( x - 3)^2 = 0
x - 3 = 0
x = 3
=====
b)
log 3x + 5 [ 30 x ] = 2 ; założenia : x > 0 ; 3x + 5 > 0; 3x + 5 różne od 1
więc
(3x + 5)^2 = 30 x
9 x^2 + 30 x + 25 = 30 x
9 x^2 = - 25 sprzeczność - brak rozwiązań [ x^2 > = 0 ]