Question

n este solutua fractiei
${3}^{(2n + 2)} \times {4}^{(2n + 3)} - 12 \times {2}^{2n} \times {6}^{2(n + 1)} = { {12}^{2} }^{4}$
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Answer 1

$3^{2n}*3^{2}*4^{2n}*4^{3}-12*2^{2n}*6^{2n+2}=12^{16}\\\\3^{2n}*3^{2}*4^{2n}*(2^2)^{3}-12*2^{2n}*6^{2n}*6^{2}=12^{16}\\\\3^{2n}*3^{2}*4^{2n}*2^{6}-12*36*12^{2n}=12^{16}\\\\12^{2n}*9*64-12*36*12^{2n}=12^{16}\\\\12^{2n}(9*64-12*36)=12^{16}\\12^{2n}(576-432)=12^{16}\\\\12^{2n}*144=12^{16}\\\\12^{2n}*12^{2}=12^{16}\\12^{2n+2}=12^{16}\\\\Bazele~sunt~egale=>si~exponentii~vor~fi.\\\\12=12=>2n+2=16\\2n=16-2\\2n=14\\n=\frac{14}{2}\\n=7\\S={7}$