Question

am nevoie de m n o p q r

Answer 1

$m)\:\log_2(2x+5)=\log_2(x^2+3x+3)\\ C.E. \left \{ {{2x+5\ \textgreater \ 0} \atop {x^2+3x+3\ \textgreater \ }} \right. \\ 2x+5=x^2+3x+3\\ x^2+x-2=0\\ \Delta=1+8=9\\ x_{1,2}=\frac{-1\pm3}{2}\\ x_1=-2\:verifica\:C.E\\ x_2=1\:verifica\:C.E\\ x\in\{-2, 1\}$

$n)\:\log_2(x+2)-log_2(x-5)=3\\ C.E. \left \{ {{x+2\ \textgreater \ 0} \atop {x-5\ \textgreater \ 0}} \right. \\ \log_2\frac{x+2}{x-5}=3\\ \frac{x+2}{x-5}=2^3=8\\ x+2=8(x-5)\\ x+2=8x-40\\ -7x=-42\\ x=6\:verifica\:C.E$

$o)\:1+\lg x=\lg(x+1)\\ C.E. \left \{ {{x\ \textgreater \ 0} \atop {x+1\ \textgreater \ 0}} \right. \\ \lg10+\lg x=\lg(x+1)\\ \lg(10x)=\lg(x+1)\\ 10x=x+1\\ 9x=1\\ x=\frac{1}{9}$

$p)\:\lg^2x-4\lg x+3=0\\ C.E.\:x\ \textgreater \ 0\\ Substitutie\:t=\lg x,\:t\ \textgreater \ 0\\ Ecuatia\:devine:\\ t^2-4t+3=0\Leftrightarrow t^2-t-3t+3=0\Leftrightarrow t(t-1)-3(t-1)=0\\ \Leftrightarrow(t-1)(t-3)=0\Rightarrow t=1\:sau\:t=3\\ Revenim\:la\:subtitutie:\\ \lg x=1\Leftrightarrow x=10\\ \lg x=3\Leftrightarrow x=10^3\\ x\in\{10,\:10^3\}$

$q)\:log_3^2 x-7log_3 x+12=0\\ C.E.\:x\ \textgreater \ 0 \\ Substitutie\:t=\log_3 x,\:t\ \textgreater \ 0\\ Ecuatia\:devine:\\ t^2-7t+12=0\\ \Delta=49-48=1\\ t_{1,2}=\frac{7\pm1}{2}\\ t_1=4\\ t_2=3 \\ Revenim\:la\:subtitutie:\\ \log_3 x=4\Leftrightarrow x=3^4\\ \log_3 x=3\Leftrightarrow x=3^3\\ x\in\{3^3,\:3^4\} [tex]r)\:\lg^2x=3\lg x-2\\ \lg^2x-3\lg x+2=0\\ C.E.\:x\ \textgreater \ 0 \\ Substitutie\:t=\lg x,\:t\ \textgreater \ 0\\ Ecuatia\:devine:\\ t^2-3t+2=0\\ \Delta=9-8=1\\ t_{1,2}=\frac{3\pm1}{2}\\ t_1=2\\ t_2=1 \\ Revenim\:la\:subtitutie:\\ \lg x=2\Leftrightarrow x=10^2\\ \lg x=1\Leftrightarrow x=10\\ x\in\{10,\:10^2\}$