Log3²x -7log3x + 5 = 0

Question

Log3²x -7log3x + 5 = 0

Answer 1

Log²(3)x-7log(3)x+5=0
x>0
log(3)x=t
t²-7t+5=0
D=49-20=29>0
t1=(7+√29)/2
t2=(7-√29)/2
1)log(3)x=(7+√29)/2
x=3^(7+√29)/2
2)log(3)x=(7-√29)/2
x=3^(7-√29)/2