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Fie \( n\in N,n\ge 2 \). Rezolvati ecuatia:
\( \log_n (x+n(n-1))=1+\sqrt[n]{\log_n x} \).

\( \log_{n}(x+n(n-1))\geq \log_{n}(n\sqrt[n]{x\cdot n^{n-1}})=\1+\frac{\log_{n}(x)+n-1}{n}\geq 1+\sqrt[n]{\log_{n}(x)} \), deci egalitate in toate inegalitatile. Rezulta ca \( x=n \) este unica solutie a ecuatiei.