Question

Log (-x^2 +2x +7)= 1+ Log (3x-7), so logaritmos estão na base 2

Answer 1

$\log_{2}-x^2+2x+7=1+\log_{2}3x-7$

$\log_{2}-x^2+2x+7-\log_{2}3x-7=1$

$\log_{2}\left(\dfrac{-x^2+2x+7}{3x-7}\right)=1$

$\dfrac{-x^2+2x+7}{3x-7}=2^1$

$-x^2+2x+7=(3x-7)\cdot2 \iff -x^2+2x+7=6x-14 \iff x^2+4x-21=0$

$\Delta=4^2-4\cdot1\cdot(-21)=16+84=100$

$x=\dfrac{-4\pm\sqrt{100}}{2}=\dfrac{-4\pm10}{2}=-2\pm5$

$x'=-2+5 \iff \boxed{x'=3}$

$x"=-2-5 \iff \boxed{x"=-7}$ (não é valida pois $3x-7$ deve ser maior que zero)