It is known that the equation loga (x-3) = 1 + loga (x + 2) + loga (x-1) about X has real roots, and the value range of real number a is the bottom I solve it as a > = (7 + 2 √ 10) / 9 or a

Simplification: x-3 = a (x + 2) (x-1)
ax*x+(a-1)x+3-2a=0
Discriminant = (A-1) (A-1) - 4a (3-2a) greater than or equal to 0
Because a is the bottom, a is greater than 0 and a is not equal to 1
My answer is yes, because a is the base, the base must be greater than 0 and not equal to 1

It is known that the equation loga (x-3) = 1 + loga (x + 2) + loga (x-1) about X has real roots, and the value range of real number a is the bottom I solve it as a > = (7 + 2 √ 10) / 9 or a

It is known that the equation loga (x-3) = 1 + loga (x + 2) + loga (x-1) about X has real roots, and the value range of real number a is the bottom I solve it as a > = (7 + 2 √ 10) / 9 or a

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