Log₉(7x+12)=log₃(7-x) OДЗ: 7x+12>0 I x>-12/7
(7-x) >0 ⇔ I x< 7 x∈(-12/7,7)
log₉(7x+12)=log₃(7-x) ⇔ (1/2)log₃(7x+12)=log₃(7-x) ⇔
log₃(7x+12)=2log₃(7-x) ⇔ log₃(7x+12)=log₃[(7-x)²]⇔
7x+12=(7-x)² ⇔x² -14x+49=7x+12⇔
x² -21x+37=0 D=21²-4·37=441-148=293
x1=21-√293 (≈21-17,12≈3,88 )∈ (-12/7,7)
x2= 21-√293(≈21+17,12≈33,17 )∉ (-12/7,7)
ответ: x1=21-√293 (≈21-17,12≈3,88 )∈ (-12/7,7)