If 5^{x-1}=(2.5)^{\log_{10}5}, then what is the value of x?

Explanation

Rewrite as 5^{x-1} = (5/2)^{\log_{10}5}. Taking \log_{10} on both sides: (x-1)\log_{10}5 = \log_{10}5 (\log_{10}5 - \log_{10}2). Divide by \log_{10}5, so x - 1 = \log_{10}5 - \log_{10}2. Since \log_{10}2 = 1 - \log_{10}5, substituting gives x - 1 = 2\log_{10}5 - 1 \implies x = 2\log_{10}5.

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