Question

14. al Determinati numerele naturale
n, pentru care
${8}^{n} + {8}^{n + 1} = 18 \times {2}^{2003}$
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Answer 1

8^n + 8^(n+1) = 18 * 2^2003 <=>
8^n * 1 + 8^n * 8 = 18 * 2^2003 <=>
8^n ( 1 + 8 ) = 18 * 2^2003 <=>
8^n * 9 = 18 * 2^2003 || : 9 <=>
8^n = 2 * 2^2003 <=>
8^n = 2^2004 <=>
8^n = 8^668 => n = 668.