Stim ca
[tex]\log_{a}{b}=\frac{1}{\log_{b}{a}}[/tex]
daca inlocuim pe a=x si b=2
[tex]\log_{x}{2}=\frac{1}{\log_{2}{x}}[/tex]
Mai stim ca
[tex]\log_{a^{2}}{b}=\frac{1}{2}\log_{a}{b}[/tex]
Atunci pentru a=2
[tex]log_{2^{2}}{b}=\frac{1}{2}\log_{2}{b}[/tex]
Avem atunci
[tex]1+2*\log_{4}{10-x}\frac{1}{\log_{2}{x}}=\frac{2}{\log_{4}{x}}\Rightarrow 1+2*\frac{1}{2}\log_{2}((10-x)}\frac{1}{\log_{2}{x}}=\frac{2}{\frac{1}{2}\log_{2}{x}}\Rightarrow 1+\frac{log_{2}((10-x))}{\log_{2}{x}}=\frac{4}{\log_{2}{x}}\Rightarrow \log_2{x}+log_{2}{(10-x)}=4\Rightarrow \log_{2}{x(10-x)}=4\Rightarrow x(10-x)=4^{2}=16\Rightarrow 10x-x^{2}=16\Rightarrow x^{2}-10x+16=0\Rightarrow x^{2}-8x-2x+16=x(x-8)-2(x-8)=(x-2)(x-8)=0[/tex]
Avem atunci solutiile:
x=2 si x=8
