Help me solve this math problem pls?

Suppose [ 5 ] denote base 5

log [5] (3x-11) + log [5] (x-27) = 3 + log [5] 8 = log [5] 5^3 + log [5] 8
log [5] (3x-11)(x-27) = log [5] (5^3)*8
(3x-11)(x-27) = (5^3)*8
3x^2 - 92x + 297 = 1000
3x^2 - 92x - 703 = 0
(x-37)(3x+19) = 0

x = 37
x = - 19/3 , rejected , since x-27 > 0

Ans: x = 37

Log[5](3x - 11) + Log[5](x - 27) = 3 + Log[5](8)

(3x - 11) > 0 → 3x - 11 > 0 → 3x > 11 → x > 11/3

(x - 27) > 0 → x - 27 > 0 → x > 27

You join the 2 conditions and you can get only one: x > 27


Log[5](3x - 11) + Log[5](x - 27) = 3 + Log[5](8) → recall: Log[a](x) = Ln(x)/Ln(a) where a is the base

[Ln(3x - 11) / Ln(5)] + [Ln(x - 27) / Ln(5)] = 3 + [Ln(8) / Ln(5)]

[Ln(3x - 11) / Ln(5)] + [Ln(x - 27) / Ln(5)] = [3.Ln(5) / Ln(5)] + [Ln(8) / Ln(5)]

Ln(3x - 11) + Ln(x - 27) = 3.Ln(5) + Ln(8) → recall: a.Ln(x) = Ln(x^a)

Ln(3x - 11) + Ln(x - 27) = Ln(125) + Ln(8) → recall: Ln(a) + Ln(b) = Ln(ab)

Ln[(3x - 11).(x - 27)] = Ln(1000)

(3x - 11).(x - 27) = 1000

3x² - 81x - 11x + 297 = 1000

3x² - 92x - 703 = 0

3.[x² - (92/3).x - (703/3)] = 0

x² - (92/3).x - (703/3) = 0

x² - (92/3).x + (46/3)² - (46/3)² - (703/3) = 0

x² - (92/3).x + (46/3)² - (2116/9) - (2109/9) = 0

x² - (92/3).x + (46/3)² - (4225/9) = 0

x² - (92/3).x + (46/3)² = 4225/9

[x - (46/3)]² = (± 65/3)²

x - (46/3) = ± 65/3

x = (46/3) ± (65/3)

x = (46 ± 65)/3

x₁ = (46 - 65)/3 → x₁ = - 19/3 ← no possible because the condition

x₂ = (46 + 65)/3 → x₂ = 37

log5 [(3x -11)(x -27) /8]=3
(3x -11)(x -27) /8 = 125
3x² -92x +297=1000
3x² -92x -703=0
x= -19/3, 37