Question

Log2(x)+log2(x-1)=log2(6)

Answer 1

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condição: x >0

$\log_2x+\log_2(x-1)=\log_26 \\ \\ aplicando ~~a ~~propriedade \\ \\ \log_2x(x-1)=\log_26 \\ \\ x(x-1)=6 \\ x^2-x-6=0 \\ \\ a=1 \\ b=-1 \\ c=-6 \\ \\ \Delta=b^2-4ac \\ \Delta=(-1)^2-4(1)(-6) \\ \Delta=1+24 \\ \Delta=25 \\ \\ x= \frac{-b\pm \sqrt{\Delta} }{2a} =~~ \frac{-(-1)\pm \sqrt{25} }{2} =~~ \frac{1\pm5}{2} \\ \\ x'= \frac{1+5}{2} = \frac{6}{2} =3 \\ \\ x"= \frac{1-5}{2} =- \frac{4}{2} =-2~~n/~~serve \\ \\ S=\{3\}$