Find the slope of the given line.

Step 1: Find the slope of the given line. The equation of the given line is $3y = 5 - 2x$. To find its slope, rewrite it in the slope-intercept form $y = mx + c$: $$ 3y = -2x + 5 $$ $$ y = -\frac{2}{3}x + \frac{5}{3} $$ The slope of this line, let's call it $m_1$, is $-\frac{2}{3}$. Step 2: Find the slope of the line passing through A and B. The line passing through points A $(2, 1)$ and B $(4, m)$ is perpendicular to the given line. If two lines are perpendicular, the product of their slopes is $-1$. Let $m_2$ be the slope of the line through A and B. $$ m_1 \times m_2 = -1 $$ $$ -\frac{2}{3} \times m_2 = -1 $$ $$ m_2 = \frac{-1}{-\frac{2}{3}} $$ $$ m_2 = \frac{3}{2} $$ Step 3: Use the slope formula to find the value of $m$. The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$. Using points A $(2, 1)$ and B $(4, m)$: $$ m_2 = \frac{m - 1}{4 - 2} $$ $$ \frac{3}{2} = \frac{m - 1}{2} $$ Multiply both sides by 2: $$ 3 = m - 1 $$ Add 1 to both sides: $$ m = 3 + 1 $$ $$ m = 4 $$ The value of $m$ is $\boxed{4}$. Step 1: Divide the equation by the coefficient of $x^2$. The given quadratic equation is $2x^2 - 5x + 3 = 0$. Divide all terms by 2: $$ x^2 - \frac{5}{2}x + \frac{3}{2} = 0 $$ Step 2: Move the constant term to the right side. $$ x^2 - \frac{5}{2}x = -\frac{3}{2} $$ Step 3: Complete the square on the left side. To complete the square for $x^2 + bx$, add $(\frac{b}{2})^2$ to both sides. Here, $b = -\frac{5}{2}$. $$ \left(\frac{b}{2}\right)^2 = \left(\frac{-\frac{5}{2}}{2}\right)^2 = \left(-\frac{5}{4}\right)^2 = \frac{25}{16} $$ Add $\frac{25}{16}$ to both sides of the equation: $$ x^2 - \frac{5}{2}x + \frac{25}{16} = -\frac{3}{2} + \frac{25}{16} $$ Step 4: Factor the left side and simplify the right side. The left side is a perfect square: $(x - \frac{5}{4})^2$. For the right side, find a common denominator: $$ -\frac{3}{2} + \frac{25}{16} = -\frac{3 \times 8}{2 \times 8} + \frac{25}{16} = -\frac{24}{16} + \frac{25}{16} = \frac{1}{16} $$ So the equation becomes: $$ \left(x - \frac{5}{4}\right)^2 = \frac{1}{16} $$ Step 5: Take the square root of both sides. $$ x - \frac{5}{4} = \pm\sqrt{\frac{1}{16}} $$ $$ x - \frac{5}{4} = \pm\frac{1}{4} $$ Step 6: Solve for $x$. $$ x = \frac{5}{4} \pm \frac{1}{4} $$ Two possible solutions: $$ x_1 = \frac{5}{4} + \frac{1}{4} = \frac{6}{4} = \frac{3}{2} $$ $$ x_2 = \frac{5}{4} - \frac{1}{4} = \frac{4}{4} = 1 $$ The solutions are $\boxed{x = \frac{3}{2} \text{ or } x = 1}$. Step 1: Apply logarithm properties to simplify the equation. The given equation is $\text{Log } (x-1) = \text{log } 12 - \text{log } (x-2)$. Use the logarithm property $\text{log } a - \text{log } b = \text{log } \left(\frac{a}{b}\right)$ on the right side: $$ \text{Log } (x-1) = \text{log } \left(\frac{12}{x-2}\right) $$ Step 2: Equate the arguments of the logarithms. Since the logarithms on both sides have the same base (implied base 10), their arguments must be equal: $$ x-1 = \frac{12}{x-2} $$ Step 3: Solve the resulting algebraic equation. Multiply both sides by $(x-2)$ to eliminate the denominator: $$ (x-1)(x-2) = 12 $$ Expand the left side: $$ x^2 - 2x - x + 2 = 12 $$ $$ x^2 - 3x + 2 = 12 $$ Subtract 12 from both sides to set the quadratic equation to zero: $$ x^2 - 3x - 10 = 0 $$ Factor the quadratic equation: We need two numbers that multiply to -10 and add to -3. These numbers are -5 and 2. $$ (x-5)(x+2) = 0 $$ This gives two possible solutions for $x$: $$ x-5 = 0 \implies x = 5 $$ $$ x+2 = 0 \implies x = -2 $$ Step 4: Check for valid solutions. For a logarithm $\text{log } A$ to be defined, the argument $A$ must be positive ($A > 0$). Check $x=5$: $x-1 = 5-1 = 4 > 0$ (Valid) $x-2 = 5-2 = 3 > 0$ (Valid) So, $x=5$ is a valid solution. Check $x=-2$: $x-1 = -2-1 = -3$ (Not valid, as argument must be positive) $x-2 = -2-2 = -4$ (Not valid) So, $x=-2$ is not a valid solution. The only valid solution is $\boxed{x=5}$. 3 done, 2 left today. You're making progress.