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Therefore e^y = (x+1)² (shown)

e^y+e^(y/2) = 6x → (1)

Sub e^y = (x+1)² into (1)

(x+1)²+[(x+1)²]^½ = 6x

x²+2x+1+x+1 = 6x

x²–3x+2 = 0

(x–2)(x–1) = 0

x–2 = 0 or x–1 = 0

x = 2 or x = 1 (rej since x > 1)

ln[(e^y)/(x²–1)] = ln(x+1)–ln(x–1) = ln[(x+1)²/(x²–1)] → (2)

Sub x = 2 into (2)

ln[(e^y)/(2²–1)] = ln[(2+1)²/(2²–1)]

ln[(e^y)/3] = ln[9/3]

e^y = 9

ln(e^y) = ln9

y = ln9