Надо доказать
㏒2 ([tex] \frac{x+3}{x+4} [/tex])=1/ ㏒2 [tex] \frac{x+4}{x+3} [/tex]
㏒2 ([tex] \frac{x+3}{x+4} [/tex])=1/ ㏒2 [tex] \frac{x+4}{x+3} [/tex]
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формула
log(2)(x+3)/(x+4)=1/log(2))(x+4)/(x+3)
log(2)(x+3/x+4)•(-log(2)(x+3/x+4))=1
-(log(2)(x+3/x+4))^2=1
(log(2)(x+3)/x+4))^2=-1
это не правильно???