Question

Rezolvati ecuatiile:
a)4^n!=64*2^n
b)(2^n)!=(n^2)!

Answer 1

a)[tex] 4^{n!}=64*2^n,rezulta, (2^2)^{n!}=2^6*2^n,sau, 2^{2(n!)}= 2^{n+6},deci, [/tex], decise obtine ecuatia 2(n!)=n+6, care nu are radacin, si anume pentru 
n=2 avem 2*1*2*<2+6, iar pentr n=3 avem: 2*1*2*3>3+6, de aici in sus inegalitatea e si mai evidenta.
b)$(2^{n})!=(n^2)!,adica,2^n=n^2,adevarata,pentu, n=2; (2^2=2^2),si,pentru,n=4 ,2^4=4^2.$, deci n∈{2; 4}.