Question

$log_{x+1}( 2x^{3} }+ 2x^{2}-3x+1)=3$

Answer 1

$\displaystyle \mathtt{log_{x+1}\left(2x^3+2x^2-3x+1\right)=3~~~~~~~~~~~~~~~~~~ C.E. \left\{\begin{array}{ccc}\mathtt{x+1\ \textgreater \ 0}\\\mathtt{2x^3+2x^2-3x+1\ \textgreater \ 0}\\\mathtt{x+1 \not = 1}\end{array}\right } \\ \\ \mathtt{log_{x+1}\left(2x^3+2x^2-3x+1 \right)=log_{x+1}(x+1)^3}\\ \\ \mathtt{2x^3+2x^2-3x+1=(x+1)^3}\\ \\ \mathtt{2x^3+2x^2-3x+1=x^3+1^3+3 \cdot x \cdot 1(x+1)}\\ \\ \mathtt{2x^3+2x^2-3x+1=x^3+1+3x(x+1)}\\ \\ \mathtt{2x^3+2x^2-3x+1=x^3+1+3x^2+3x}\\ \\ \mathtt{2x^3+2x^2-3x+1-x^3-1-3x^2-3x=0}$
$\displaystyle \mathtt{x^3-x^2-6x=0} \\ \\ \mathtt{x(x+2)(x-3)=0} \\ \\ \mathtt{x=0;~~~~~~~~~~~~~~x+2=0 \Rightarrow x=-2;~~~~~~~~~~~~~~x-3=0 \Rightarrow x=3}\\ \\ \mathtt{x=0 \Rightarrow \left\{\begin{array}{ccc}\mathtt{0+1\ \textgreater \ 0~A}\\\mathtt{2 \cdot 0^3+2 \cdot 0^2-3 \cdot 0+1\ \textgreater \ 0~A}\\ \mathtt{0+1 \not= 1~F}\end{array}\right \Rightarrow x=0~nu~este~solutie}\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mathtt{a~ecuatiei.}$

$\displaystyle \mathtt{x=-2 \Rightarrow -2+1\ \textgreater \ 0 ~F \Rightarrow x=-2~nu~este~solutie~a~ecuatiei.} \\ \\ \mathtt{x=3 \Rightarrow \left\{\begin{array}{ccc}\mathtt{3+1\ \textgreater \ 0~A}\\\mathtt{2 \cdot 3^3+2 \cdot 3^2-3 \cdot 2+1\ \textgreater \ 0~A}\\3+1 \not = 1~A\end{array}\right \Rightarrow x=3~este~solutie~a}\\ \mathtt{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ecuatiei.}\\ \\ \mathtt{S=\{3\}}$

Answer 2

$2x^3+2 x^{2} -3x+1=(x+1)^3 \\ 2x^3+2 x^{2} -3x+1=x^3+3 x^{2} +3x+1 \\ x^3- x^{2} -6x=0 \\ x( x^{2} -x-6)=0 \\ x=0;x=-2;x=3$
luind in cosideratie ca baza unui logaritm este pozitiva si diferita de 1,rezulta ca
S={3}