Question

Aflati numerele naturale n si k stiind ca 2^n×3^k-2^n+3=2n+1

Answer 1

[tex]2^n(3^k-1)=2n-2\\ 2^n(3^k-1)=2(n-1)\\ 2^{n-1}(3^k-1)=n-1\\ n=1,k=0[/tex]
solutia este unica deoarece ptr. n>1, $2^{n-1}\ \textgreater \ n-1$