Solve the equation \( \log _{10}(4 x+1)-\log _{10}(3 x-2)=\log _{10}(2 x-1)-\log _{10}(2 x-3) \)

To solve the equation \[ \log_{10}(4x + 1) - \log_{10}(3x - 2) = \log_{10}(2x - 1) - \log_{10}(2x - 3), \] we can use the properties of logarithms. Specifically, we can rewrite the differences of logs as ratios: \[ \log_{10}\left(\frac{4x + 1}{3x - 2}\right) = \log_{10}\left(\frac{2x - 1}{2x - 3}\right). \] Since the logarithm function is one-to-one, we can set the arguments equal to each other: \[ \frac{4x + 1}{3x - 2} = \frac{2x - 1}{2x - 3}. \] Cross-multiplying gives us: \[ (4x + 1)(2x - 3) = (2x - 1)(3x - 2). \] Expanding both sides: \[ 8x^2 - 12x + 2x - 3 = 6x^2 - 4x - 3x + 2. \] This simplifies to: \[ 8x^2 - 10x - 3 = 6x^2 - 7x + 2. \] Combining like terms results in: \[ 8x^2 - 6x^2 - 10x + 7x - 3 - 2 = 0, \] which simplifies to: \[ 2x^2 - 3x - 5 = 0. \] Now we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), and \(c = -5\): \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4}. \] This gives us two potential solutions: 1. \(x = \frac{10}{4} = 2.5\) 2. \(x = \frac{-4}{4} = -1\) Now we need to verify that these values satisfy the original logarithmic constraints: For \(x = 2.5\): - \(4x + 1 = 4(2.5) + 1 = 10 + 1 = 11\) (valid) - \(3x - 2 = 3(2.5) - 2 = 7.5 - 2 = 5.5\) (valid) - \(2x - 1 = 5 - 1 = 4\) (valid) - \(2x - 3 = 5 - 3 = 2\) (valid) For \(x = -1\): - \(4x + 1 = 4(-1) + 1 = -4 + 1 = -3\) (invalid, as logs of negative numbers are undefined) Thus, the only valid solution is \[ \boxed{2.5}. \]