Question

Como resolve essa equação log x (na base 2) + log (x -1) = 1 (na base 2).

Answer 1

$log_2 x + \log_2(x -1) = 1\\\\\\Condic\~ao~de~existe\^encia\\\\x-1\ \textgreater \ 0\\x\ \textgreater \ 1\\\\\\\log_2x+\log_2(x-1)=\log_22\\\log_2(x.[x-1])=\log_22\\\log_2(x^2-x)=\log_22\\\\x^2-x=2\\x^2-x-2=0\\\\(x-2).(x+1)=x^2-x-2\\\\\mathsf{S=(2)}$

Dúvidas? comente