Решить уравнение и найти корни принадлежащие отрезку

$3x - log_{6} 2^{3x} = log_{6}( 3^{3x} + x^{2} - 9 )$
3x - 3x · log₆2 = log₆(3³ˣ + x² - 9)
3x(1 - log₆2) = log₆(3³ˣ + x² - 9)
3x(1 - 1/log₂6) = log₆(3³ˣ + x² - 9)
3x(1 - 1/ (1 + log₂3)) = log₆(3³ˣ + x² - 9)
3x ( (1 + log₂3 - 1) / (1 + log₂3)) = log₆(3³ˣ + x² - 9)
3x · log₂3 / (1 + log₂3) = log₂ (3³ˣ + x² - 9) / log₂ 6
3x · log₂3 / (1 + log₂3) = log₂ (3³ˣ + x² - 9) / (1 + log₂3)
log₂ 3³ˣ = log₂ (3³ˣ + x² - 9)
3³ˣ = 3³ˣ + x² - 9
x² - 9 = 0
x = - 3
x = 3