[tex]\displaystyle \mathtt{IV.~1)log_ \frac{1}{5}\left( \frac{x+2}{10}\right)=log_ \frac{1}{5} \left( \frac{2}{x+1}\right) ~~~~~~~~~~~~~~~~~~~~~~C.E. \left \{ {{ \frac{x+2}{10}\ \textgreater \ 0 } \atop { \frac{2}{x+1}\ \textgreater \ 0 }} \right. }\\ \\ \mathtt{\frac{x+2}{10} = \frac{2}{x+1} } \\ \\ \mathtt{(x+2)(x+1)=10 \cdot 2}\\ \\ \mathtt{x^2+x+2x+2=20}\\ \\ \mathtt{x^2+3x+2-20=0}\\ \\ \mathtt{x^2+3x-18=0}\\ \\ \mathtt{a=1,~b=3,~c=-18}\\ \\ \mathtt{\Delta=b^2-4ac=3^2-4 \cdot 1 \cdot (-18)=9+72=81\ \textgreater \ 0}[/tex]
[tex]\displaystyle \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-3+ \sqrt{81} }{2 \cdot 1}= \frac{-3+9}{2}= \frac{6}{2} =3}\\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-3- \sqrt{81} }{2 \cdot 1} = \frac{-3-9}{2} = \frac{-12}{2}=-6} [/tex]
[tex]\displaystyle \mathtt{x_1=3 \Rightarrow \left \{ {{ \frac{3+2}{10}\ \textgreater \ 0~A } \atop { \frac{2}{3+1}\ \textgreater \ 0~A }} \right. \Rightarrow x_1=3~este~solutie~a~ecuatiei} \\ \\ \mathtt{x_2=-6 \Rightarrow \left \{ {{ \frac{-6+2}{10}\ \textgreater \ 0~F } \atop { \frac{2}{-6+1}\ \textgreater \ 0~F }} \right. \Rightarrow x_2=-6~nu~este~solutie~a~ecuatiei } \\ \\ S=\{3\}[/tex]
[tex]\displaystyle \mathtt{2)log_{ \frac{1}{3} }\left(x^2+3x-4\right)=log_{ \frac{1}{3} }(2x+2)~~~~~~~~~~~~~~~~C.E. \left \{ {{x^2+3x-4\ \textgreater \ 0} \atop {2x+2\ \textgreater \ 0}} \right. }\\ \\ \mathtt{x^2+3x-4=2x+2}\\ \\ \mathtt{x^2+3x-4-2x-2=0}\\ \\ \mathtt{x^2+x-6=0}\\ \\ \mathtt{a=1,~b=1,~c=-6}\\ \\ \mathtt{\Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-6)=1+24=25\ \textgreater \ 0}[/tex]
[tex]\displaystyle \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-1+ \sqrt{25} }{2 \cdot 1}= \frac{-1+5}{2} = \frac{4}{2}=2}\\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-1- \sqrt{25} }{2 \cdot 1}= \frac{-1-5}{2}= \frac{-6}{2}=-3 } [/tex]
[tex]\displaystyle \mathtt{x_1=2 \Rightarrow \left \{ {{2^2+3 \cdot 2-4\ \textgreater \ 0~A} \atop {2 \cdot 2+2\ \textgreater \ 0 ~A}} \right. \Rightarrow x_1=2 ~este~solutie~a~ecuatiei}\\ \\ \mathtt{x_2=-3 \Rightarrow \left \{ {{(-3)^2+3 \cdot (-3)-4\ \textgreater \ 0~F} \atop {2 \cdot (-3)+2\ \textgreater \ 0~F}} \right. \Rightarrow x_2=-3~nu~este~solutie}\\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a~ecuatiei} \\ \\ \mathtt{S=\{2\}}[/tex]
[tex]\displaystyle \mathtt{3)log_4\left(3^x+1\right)=log_4 \left(-2^x+14\right)~~~~~~~~~~~~~~~~~~~~C.E. \left \{ {{3^x+1\ \textgreater \ 0} \atop {-2^x+14\ \textgreater \ 0}} \right. }\\ \\ \mathtt{3^x+1=-2^x+14}\\ \\ \mathtt{3^x+1+2^x-14=0} \\ \\ \mathtt{3^x+2^x-13=0}\\ \\ 3^x+2^x=13 \\ \\ \mathtt{x=2} \\ \\ \mathtt{x=2 \Rightarrow \left \{ {{3^2+1\ \textgreater \ 0~A} \atop {-2^2+14\ \textgreater \ 0~A}} \right. \Rightarrow x=2~este~solutie~a~ecuatiei }[/tex]
[tex]\displaystyle \mathtt{S=\{2\}}[/tex]
[tex]\displaystyle \mathtt{4)log_{x-2}(2x-9)=log_{x-2}(23-6x)}~~~~~~~~~~~~~~~~~~~ \mathtt{~C.E. \left\{\begin{array}{ccc}\mathtt{x-2\ \textgreater \ 0}\\\mathtt{2x-9\ \textgreater \ 0}\\\mathtt{23-6x\ \textgreater \ 0}\\\mathtt{x-2 \not =1}\end{array}\right } \\ \\ \mathtt{2x-9=23-6x } \\ \\ \mathtt{2x+6x=23+9}\\ \\ \mathtt{8x=32}\\ \\ \mathtt{x=4} \\ \\ \mathtt{x=4\Rightarrow \left\{\begin{array}{ccc}\mathtt{4-2\ \textgreater \ 0~A}\\\mathtt{2 \cdot 4-9\ \textgreater \ 0~F}\\\mathtt{23-6 \cdot 4\ \textgreater \ 0~F}\\\mathtt{4-2 \not =1~A}\end{array}\right \Rightarrow x=4~nu~este~solutie~a~ecuatiei}[/tex]
[tex]\displaystyle \mathtt{5)log_{5x-2}\left(2x^2\right)=log_{5x-2}(x+1)~~~~~~~~~~~~~~~~~~~~ C.E.\left\{\begin{array}{ccc}\mathtt{5x-2\ \textgreater \ 0}\\\mathtt{2x^2\ \textgreater \ 0}\\\mathtt{x+1\ \textgreater \ 0}\\\mathtt{5x-1 \not = 1}\end{array}\right}~\\ \\ \mathtt{2x^2=x+1 }\\ \\ \mathtt{2x^2-x-1=0}\\ \\ \mathtt{a=2,~b=-1,~c=-1}\\ \\ \mathtt{\Delta=b^2-4ac=(-1)^2-4 \cdot 2 \cdot (-1)=1+8=9\ \textgreater \ 0}[/tex]
[tex]\displaystyle \mathtt{x_1= \frac{-b+ \sqrt{\Delta} }{2a}= \frac{-(-1)+ \sqrt{9} }{2 \cdot 2}= \frac{1+3}{4} = \frac{4}{4}=1}\\ \\ \mathtt{x_2= \frac{-b- \sqrt{\Delta} }{2a}= \frac{-(-1)- \sqrt{9} }{2 \cdot 2}= \frac{1-3}{4}= \frac{-2}{4}=- \frac{1}{2} } \\ \\ \mathtt{x=1 \Rightarrow \left\{\begin{array}{ccc}\mathtt{5 \cdot 1-2\ \textgreater \ 0~A}\\\mathtt{2 \cdot 1^2\ \textgreater \ 0~A}\\\mathtt{1+1\ \textgreater \ 0~A}\\\mathtt{5 \cdot 1-2 \not = 0~A}\end{array}\right \Rightarrow x=1~este~solutie~a~ecuatiei}[/tex]
[tex]\displaystyle \mathtt{x=- \frac{1}{2}\Rightarrow \left\{\begin{array}{ccc}\mathtt{5 \cdot \left(-\frac{1}{2}\right)-2\ \textgreater \ 0~F }\\\mathtt{2 \cdot \left(- \frac{1}{2}\right)^2\ \textgreater \ 0~A }\\\mathtt{- \frac{1}{2}+1\ \textgreater \ 0~A }\\\mathtt{5 \cdot \left(- \frac{1}{2} \right)-2 \not= 1~A}\end{array}\right \Rightarrow x=- \frac{1}{2}~nu~este~solutie~a~ecuatiei}\\ \\ S=\{1\}[/tex]
