Question

lg(x-2)- 1/2*lg(4x-8)=lg2 URGEEEEEENT!

Answer 1

$\lg(x-2)- \dfrac{1}{2}\cdot\lg(4x-8)=\lg2 \\ \\ \lg(x-2) -\dfrac{1}{2}\cdot \lg\Big(4(x-2)\Big) = \lg 2 \\ \\ \lg(x-2) - \dfrac{1}{2}\cdot \Big(\lg4+\lg(x-2)\Big) = \lg 2\\ \\ \lg(x-2) -\dfrac{1}{2}\cdot \lg 4-\dfrac{1}{2}\cdot \lg(x-2) = \lg 2 \\ \\ Notam \lg(x-2) = t, \quad t\ \textgreater \ 0.\\ \\ t-\lg4^\dfrac{1}{2}} - \dfrac{1}{2}\cdot t = \lg 2\Big|\cdot 2 \\ \\ 2t - 2\lg \sqrt4-t = 2\lg 2 \\ \\ t - 2\lg 2 = 2\lg 2\\ \\ t - \lg2^2 = \lg2^2 \\ \\ t-\lg4 = \lg4 \\ \\ t = 2\lg4 \\ \\ t = \lg4^2 \\ \\ t = \lg16$

$\\ Revenim la notatie: \\ \\ \lg(x-2) = \lg16 \\ \\ Conditie de existenta: x-2 \ \textgreater \ 0 \Rightarrow x\ \textgreater \ 2 \Rightarrow D = (2,+\infty)\\ \\ \lg(x-2) = \lg16\\ \\ x-2 = 16 \\ \\ x = 18\in D\\ \\ \Rightarrow \boxed{S = \Big\{18\Big\}}$