Question

$log_{2}( x^{2}-2)- log_{2}(x+ \frac{1}{2} )=1$

Answer 1

$x^{2} -2=2(x+ \frac{1}{2}) \\ x^{2} -2=2x+1 \\ x^{2} -2x-3=0 \\ x_{1}=-1; x_{2}=3$
x=-1 nu indeplineste conditia logaritmului,deci
S={3}

Answer 2

$\displaystyle \mathtt{log_2\left(x^2-2\right)-log_2 \left(x+ \frac{1}{2}\right) =1~~~~~~~~~~~~~~~~~~~~~~~~~~ C.E. \left \{ {{x^2-2\ \textgreater \ 0} \atop {x+ \frac{1}{2}\ \textgreater \ 0 }} \right. }\\ \\ \mathtt{log_2\left(x^2-2\right)=1+log_2 \left(x+ \frac{1}{2}\right) }\\ \\ \mathtt{log_2\left(x^2-2\right)=log_22+log_2 \left(x+ \frac{1}{2}\right) }\\ \\ \mathtt{log_2\left(x^2-2\right)=log_2 \left(2\left(x+ \frac{1}{2}\right)\right)}\\ \\ \mathtt{x^2-2=2 \left(x+ \frac{1}{2}\right) }$
$\displaystyle \mathtt{x^2-2=2x+1}\\ \\ \mathtt{x^2-2-2x-1=0} \\ \\ \mathtt{x^2-2x-3=0}\\ \\ \mathtt{a=1,~ b=-2,~c=-3 }\\ \\ \mathtt{\Delta=b^2-4ac=(-2)^2-4 \cdot 1 \cdot (-3)=4+12=16\ \textgreater \ 0}\\ \\ \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-2)+ \sqrt{16} }{2 \cdot 1} = \frac{2+4}{2}= \frac{6}{2}=3 }\\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-(-2)- \sqrt{16} }{2 \cdot 1}= \frac{2-4}{2}= \frac{-2}{2}=-1 }$
$\displaystyle \mathtt{x_1=3 \Rightarrow \left \{ {{3^2-2\ \textgreater \ 0~A} \atop {3+ \frac{1}{2}}\ \textgreater \ 0~A} \right. \Rightarrow x_1=3~este~solutie~a~ecuatiei}\\ \\ \mathtt{x_2=-1 \Rightarrow (-1)^2-2\ \textgreater \ 0~F \Rightarrow x_2=-1~nu~este~solutie~a~ecuatiei}\\ \\ \mathtt{S=\{3\}}$