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Solve 9^x + 4^x + 1 = 6^x + 3^x + 2^x | IOQM Algebra

JEE Maths question with a full step-by-step solution.

Question
Find all real numbers xx such that
9x+4x+1=6x+3x+2x.9^x + 4^x + 1 = 6^x + 3^x + 2^x.
Solution
Answer: 0
Step 1: Substitute. Let p=3xp = 3^x and q=2xq = 2^x, both positive. Then 9x=p29^x = p^2, 4x=q24^x = q^2, 6x=pq6^x = pq, 3x=p3^x = p and 2x=q2^x = q, so the equation becomes
p2+q2+1=pq+p+q.p^2 + q^2 + 1 = pq + p + q.
 Step 2: Form a sum of squares. Move all terms to one side,
p2+q2+1−pq−p−q=0,p^2 + q^2 + 1 - pq - p - q = 0,
 and multiply by 22,
2p2+2q2+2−2pq−2p−2q=0,2p^2 + 2q^2 + 2 - 2pq - 2p - 2q = 0,
 which regroups as
(p−q)2+(q−1)2+(1−p)2=0.(p - q)^2 + (q - 1)^2 + (1 - p)^2 = 0.
 Step 3: A sum of squares of real numbers is zero only when each square is zero. Hence p−q=0p - q = 0, q−1=0q - 1 = 0 and 1−p=01 - p = 0, giving p=q=1p = q = 1. Step 4: Back-substitute: 3x=13^x = 1 and 2x=12^x = 1, so x=0x = 0. Answer: x=0x = 0 is the only real solution
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