Solve the equation log_(2)(4x-3)+log_(2)(3x+5)=3

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Combine Logs: We can use the property of logarithms that allows us to combine the sum of two logs with the same base into a single log by multiplying the arguments. So, we combine log₂(4x-3) and log₂(3x+5) into a single logarithm. log₂((4x-3)(3x+5)) = 3Exponential Form: Next, we can rewrite the equation in exponential form to eliminate the logarithm. The equation log₂(y) = 3 is equivalent to 2³ = y. (4x-3)(3x+5) = 2³Calculate Right Side: Now we calculate 2³ to simplify the right side of the equation. (4x-3)(3x+5) = 8Expand Left Side: We expand the left side of the equation by multiplying the binomials. 4x * 3x + 4x * 5 - 3 * 3x - 3 * 5 = 8 12x² + 20x - 9x - 15 = 8Combine Like Terms: Combine like terms on the left side of the equation. 12x² + 11x - 15 = 8Set Equation to Zero: Subtract 8 from both sides to set the equation to zero, which is necessary for solving a quadratic equation. 12x² + 11x - 15 - 8 = 0 12x² + 11x - 23 = 0Solve Quadratic Equation: Now we need to solve the quadratic equation. This can be done by factoring, completing the square, or using the quadratic formula. The quadratic formula is x = (-b ± √(b² - 4ac)) / (2a). In this case, a = 12, b = 11, and c = -23.Calculate Discriminant: We calculate the discriminant (b² - 4ac) to determine if there are real solutions. Discriminant = 11² - 4(12)(-23) Discriminant = 121 + 1104 Discriminant = 1225Apply Quadratic Formula: Since the discriminant is positive, there are two real solutions. We proceed with the quadratic formula. x = (-11 ± √1225) / (24)Calculate Square Root: Calculate the square root of the discriminant. √1225 = 35Substitute Square Root: Substitute the square root back into the quadratic formula. x = (-11 ± 35) / (24)Solve for x: Solve for the two possible values of x. x₁ = (-11 + 35) / 24 x₁ = 24 / 24 x₁ = 1  x₂ = (-11 - 35) / 24 x₂ = -46 / 24 x₂ = -23 / 12Check Valid Solutions: We must check these solutions to ensure they do not make the argument of the original logarithms negative, as the logarithm of a negative number is undefined. For x₁ = 1: 4(1) - 3 = 4 - 3 = 1 (positive) 3(1) + 5 = 3 + 5 = 8 (positive)  For x₂ = -23/12: 4(-23/12) - 3 = -23/3 - 3 < 0 (negative) 3(-23/12) + 5 = -23/4 + 5 < 0 (negative)  x₂ = -23/12 is not a valid solution because it makes the arguments of the logarithms negative.Final Valid Solution: Therefore, the only valid solution is x₁ = 1.

Solve the equation $\log_{2}(4x-3)+\log_{2}(3x+5)=3$

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Solve the equation $\log_{2}(4x-3)+\log_{2}(3x+5)=3$