Question

aratati ca nr a=log2(6+4radical2)+ 2log2(2-radical 2) este natural
$a = log2(6 + 4 \sqrt{2) } + 2l \\ og2(2 - \sqrt{2} )$
va rog, îmi trebuie urgent​

Answer 1

Am atasat o rezolvare.

Answer 2

Răspuns:

Explicație pas cu pas:

$a=log_{2}(6+4\sqrt{2})+2*log_{2}(2-\sqrt{2})= log_{2}(6+4\sqrt{2})+log_{2}(2-\sqrt{2})^{2}=log_{2}(6+4\sqrt{2})+log_{2}(2^{2}-2*2*\sqrt{2}+(\sqrt{2})^{2} )=log_{2}(6+4\sqrt{2})+log_{2}(6-4\sqrt{2})=log_{2}[(6+4\sqrt{2})(6-4\sqrt{2})]=log_{2}[6^{2}-(4\sqrt{2})^{2}]=log_{2}[36-32]=log_{2}4=2$

2∈N