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[tex] log_{3}(2x + 1) - log_{3}(2x - 1) = - 1[/tex]

[tex]log_3(2x+1)=log_3(2x-1)-log_33 [/tex]
[tex]log_3(2x+1)=log_3 \frac{2x-1}{3} [/tex]
[tex]2x+1>0 [/tex]
[tex] \frac{2x-1}{3} \ \textgreater \ 0[/tex]
Functia logaritmica este injectiva
[tex]f(x_1)=f(x_2)==\ \textgreater \ x_1=x_2[/tex]
[tex]2x+1=\frac{2x-1}{3}[/tex]
[tex] 6x+3=2x-1[/tex]
[tex]4x=-4[/tex]
[tex]x=-1[/tex]
[tex]2x+1\ \textgreater \ 0 -1\ \textgreater \ 0 (F)[/tex]
===> S=∅

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Rayzen Rayzen · Answer 2 · 2017-12-22T16:45:13+02:00

[tex]\log_{3}(2x+1) - \log_{3}(2x-1) = -1\\ \\ $Conditii de existenta:\\ \\ \left\{ \begin{array}{ll} 2x+1\ \textgreater \ 0 \\ 2x-1\ \textgreater \ 0 \end{array} \right \Rightarrow \left\{ \begin{array}{ll} 2x\ \textgreater \ -1 \\ 2x\ \textgreater \ 1\end{array} \right \Rightarrow \left\{ \begin{array}{ll} x\ \textgreater \ -\dfrac{1}{2} \\ x\ \textgreater \ \dfrac{1}{2}\end{array} \right| \Rightarrow D = \Big(\dfrac{1}{2},+\infty \Big)\\ \\ \\ \log_\big3 (2x+1) - \log_\big3(2x-1) = -1 \\ \\\log_\big3\dfrac{2x+1}{2x-1} = -1 \\ \\ \dfrac{2x+1}{2x-1} = 3^{-1}\Big|\cdot 3 \\ \\ [/tex]

[tex]3\cdot \dfrac{2x+1}{2x-1} = 1\\ \\ \dfrac{6x+3}{2x-1} = 1 \\ \\ 6x+3 = 2x-1 \\ \\ 4x = -4 \\ \\ x = -1 \notin D \\ \\ \Rightarrow \boxed{S = \emptyset}[/tex]