Logx+log(x+3)=1 помогите пожалуйста

0\ \wedge\ x+3 > 0\\x > 0\ \wedge\ x > -3\Rightarrow x\in(-3;\ \infty)\\===============\\lgx+lg(x+3)=lg10\\lg[x(x+3)]=lg10\iff x^2+3x=10\\\\x^2+3x-10=0\\a=1;\ b=3;\ c=-10\\\Delta=b^2-4ac\to\Delta=3^2-4\cdot1\cdot(-10)=9+40=49\\\sqrt\Delta=\sqrt{49}=7\\\\x_1=\frac{-b-\sqrt\Delta}{2a}\to x_1=\frac{-3-7}{2\cdot1}=\frac{-10}{2}=-5\notin(-3;\ \infty)\\\\x_2=\frac{-b+\sqrt\Delta}{2a}\to x_2=\frac{-3+7}{2\cdot1}=\frac{4}{4}=2\in(-3;\ \infty)\\\\O:x=2." alt="lgx+lg(x+3)=1\\==============\\x > 0\ \wedge\ x+3 > 0\\x > 0\ \wedge\ x > -3\Rightarrow x\in(-3;\ \infty)\\===============\\lgx+lg(x+3)=lg10\\lg[x(x+3)]=lg10\iff x^2+3x=10\\\\x^2+3x-10=0\\a=1;\ b=3;\ c=-10\\\Delta=b^2-4ac\to\Delta=3^2-4\cdot1\cdot(-10)=9+40=49\\\sqrt\Delta=\sqrt{49}=7\\\\x_1=\frac{-b-\sqrt\Delta}{2a}\to x_1=\frac{-3-7}{2\cdot1}=\frac{-10}{2}=-5\notin(-3;\ \infty)\\\\x_2=\frac{-b+\sqrt\Delta}{2a}\to x_2=\frac{-3+7}{2\cdot1}=\frac{4}{4}=2\in(-3;\ \infty)\\\\O:x=2." align="absmiddle" class="latex-formula">

$log_ab=c\iff a^c=b\\\\loga=log_{10}a\\\\lg10=log_{10}10=1\to10^1=10$

$b=log_aa^b$