Разбираемся:
log3x+1(2x + 1) = 1 + 2log2x+1(3x + 1) => log3x+1(2x + 1) = 1 + 2log2x+1(3x + 1)
Пусть a = 3x + 1 , b = 2x + 1
log a(b) = 1 + 2log b(a)
log a(b) - 1 = 2log b(a)
(log a(b) - 1) / 2 = log b(a)