AlgebraDifficulty 5.1AIME, harderProve itUnited States
Problem: Suppose x, y, and z are real numbers greater than 1 such that xlogyz=2ylogzx=4, and zlogxy=8 Compute logxy.
Solution
Solution: Taking log2 both sides of the first equation gives log2xlogyz=1log2ylog2xlog2z=1 Performing similar manipulations on the other two equations, we get log2ylog2xlog2z=1log2zlog2ylog2x=2log2xlog2zlog2y=3 Multiplying the first and second equation gives (log2x)2=2 or log2x=±2. Multiplying the second and third equation gives (log2y)2=6 or log2y=±6. Thus, we have logxy=log2xlog2y=±26=±3
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Source: MathNet,
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