Redukowanie "a"
[tex]a=log_{\sqrt{6} }(3\sqrt{2}-2\sqrt{3})+log_{\sqrt{6}}(3\sqrt{2}+2\sqrt{3})=\\\\log_{\sqrt{6}}=[(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})]=\\\\log_{\sqrt{6} }[(3\sqrt{2} )^2-(2\sqrt{3} )^2]=log_{\sqrt{6} }[9\cdot2-4\cdot3]=\\\\log_{\sqrt{6}}[18-12]=log_{\sqrt{6}}(6)=2[/tex]
Podstawiamy pod wyrażenie za "a" - "2" i liczymy:
[tex]log_a[log_{\sqrt{2}}a]=log_2[log_{\sqrt{2}}2]=log_2[2]=1[/tex]
