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Question: (a) $log_{x+1}(x^2+x-6)^2=4$...

(a) logx+1(x2+x−6)2=4log_{x+1}(x^2+x-6)^2=4

Answer

x=1

Explanation

Solution

The domain requires x+1>0x+1 > 0, x+1≠1x+1 \neq 1, and (x2+x−6)2>0(x^2+x-6)^2 > 0. This leads to x∈(−1,0)∪(0,2)∪(2,∞)x \in (-1, 0) \cup (0, 2) \cup (2, \infty). The equation (x+1)4=(x2+x−6)2(x+1)^4 = (x^2+x-6)^2 implies (x+1)2=x2+x−6(x+1)^2 = x^2+x-6 or (x+1)2=−(x2+x−6)(x+1)^2 = -(x^2+x-6). Case 1: (x+1)2=x2+x−6  ⟹  x2+2x+1=x2+x−6  ⟹  x=−7(x+1)^2 = x^2+x-6 \implies x^2+2x+1 = x^2+x-6 \implies x = -7 (rejected). Case 2: (x+1)2=−(x2+x−6)  ⟹  x2+2x+1=−x2−x+6  ⟹  2x2+3x−5=0(x+1)^2 = -(x^2+x-6) \implies x^2+2x+1 = -x^2-x+6 \implies 2x^2+3x-5 = 0. Factoring gives (2x+5)(x−1)=0(2x+5)(x-1) = 0, so x=1x=1 or x=−5/2x=-5/2. x=1x=1 is valid. x=−5/2x=-5/2 is rejected. The only valid solution is x=1x=1.