Log2(x-3)+log2(2x+1)=2

To solve the equation ${\log }_2(x-3)+{\log }_2(2x+1)=2$, follow these steps: 1. Determine the Domain (ОДЗ) The arguments of a logarithm must be strictly positive:

• $x-3>0⟹x>3$ $2x+1>0⟹x>-0.5$

Combining these conditions, the solution must satisfy: $x>3$. 2. Apply Logarithm Properties Using the product rule ${\log }_b\left(m\right)+{\log }_b\left(n\right)={\log }_b(m\cdot n)$, we can combine the terms: ${\log }_2\left(\right(x-3\left)\right(2x+1\left)\right)=2$Rewrite the equation in exponential form ( $b^y=x$): $(x-3)(2x+1)=2^2$ $(x-3)(2x+1)=4$3. Solve the Quadratic Equation Expand the brackets: $2x^2+x-6x-3=4$ $2x^2-5x-7=0$Use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ :

• $a=2,b=-5,c=-7$ $D=(-5{)}^2-4(2\left)\right(-7)=25+56=81$ $\sqrt{D}=9$

Find the roots: $x_1=\frac{5+9}{4}=\frac{14}{4}=3.5$ $x_2=\frac{5-9}{4}=\frac{-4}{4}=-1$ 4. Check Against the Domain

• $x_1=3.5$: This value is greater than 3, so it is a valid solution. $x_2=-1$: This value is not greater than 3, so it is extraneous (invalid).

Answer: $x=3.5$ Would you like me to solve another logarithmic equation or demonstrate how to graph this function?