\log_2((x+3)(x-1)) = 3 \implies (x+3)(x-1) = 2^3 = 8

["Solving the Logarithmic Equation: How to Solve ⟨log₂((x+3)(x-1))⟩ = 3 Step-by-Step", "Understanding logarithmic equations can be a rewarding challenge in algebra. One common type involves expressing a product inside a log as a power of the base of the logarithm. In this article, we explore how to solve the equation:", "$$\n\log_2((x+3)(x-1)) = 3\n$$\nand demonstrate how it leads naturally to the equivalent expression:\n$$\n(x+3)(x-1) = 2^3 = 8\n$$", "---", "### Step 1: Understand the Logarithmic to Exponential Conversion", "The fundamental rule of logarithms states that:\n$$\n\log_b(A) = C \quad \ ext{is equivalent to} \quad A = b^C\n$$", "Applying this to our equation:", "$$\n\log_2((x+3)(x-1)) = 3 \quad \Rightarrow \quad (x+3)(x-1) = 2^3\n$$", "Since ( 2^3 = 8 ), we immediately simplify:", "$$\n(x+3)(x-1) = 8\n$$", "---", "### Step 2: Expand the Left-Hand Side", "Next, expand the product ( (x+3)(x-1) ):", "$$\n(x+3)(x-1) = x^2 - x + 3x - 3 = x^2 + 2x - 3\n$$", "So the equation becomes:", "$$\nx^2 + 2x - 3 = 8\n$$", "---", "### Step 3: Form a Standard Quadratic Equation", "Subtract 8 from both sides:", "$$\nx^2 + 2x - 3 - 8 = 0 \quad \Rightarrow \quad x^2 + 2x - 11 = 0\n$$", "---", "### Step 4: Solve the Quadratic Equation", "Use the quadratic formula:\n$$\nx = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\n$$\nFor ( x^2 + 2x - 11 = 0 ), we have ( a=1 ), ( b=2 ), ( c=-11 ).", "Calculate the discriminant:", "$$\n\Delta = 2^2 - 4(1)(-11) = 4 + 44 = 48\n$$", "So,", "$$\nx = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}\n$$", "---", "### Step 5: Check Domain Restrictions", "Remember, the original equation contains the expression ( \log_2((x+3)(x-1)) ). Logarithms are only defined when their argument is positive:", "$$\n(x+3)(x-1) > 0\n$$", "Solving ( (x+3)(x-1) > 0 ) gives:", "- Critical points at ( x = -3 ) and ( x = 1 )\n- Test intervals:\n - ( x < -3 ): positive × negative → negative\n - ( -3 < x < 1 ): negative × negative → positive\n - ( x > 1 ): positive × positive → positive", "So the solution interval is:\n$$\nx \in (-3, 1) \cup (1, \infty)\n$$", "Now check if the solutions ( x = -1 \pm 2\sqrt{3} ) fall in this domain:", "- ( 2\sqrt{3} \approx 3.464 )\n- ( x_1 = -1 + 3.464 = 2.464 ) → greater than 1: ✅ valid\n- ( x_2 = -1 - 3.464 = -4.464 ) → less than -3: ❌ invalid", "Thus, the only valid solution is ( x = -1 + 2\sqrt{3} )", "---", "### Conclusion", "The equation\n$$\n\log_2((x+3)(x-1)) = 3\n$$\nis correctly transformed by converting the logarithmic form to exponential form, yielding:\n$$\n(x+3)(x-1) = 2^3 = 8\n$$", "This elegant step connects logarithmic reasoning with algebraic manipulation. While both forms are mathematically equivalent, expressing the argument of the log as a power of 2 efficiently simplifies solving. Always remember to check the domain — an important habit when working with logarithms.", "---", "Key Takeaways:\n- Logarithmic equations convert naturally to exponential form.\n- Solving ( (x+3)(x-1) = 8 ) leads effortlessly to a quadratic.\n- Always verify solutions fit the original logarithmic expression’s domain.\n- Mastering these steps strengthens algebraic fluency and problem-solving confidence.", "---", "### Related Keywords for SEO Optimization:\n- Solve log base 2 equations\n- How to solve log equations step by step\n- Convert log to exponential form\n- Solve quadratic from logarithmic equation\n- Domain restrictions logarithmic functions\n- Logarithmic practice problems 2024", "---", "If you found this explanation helpful, share it with fellow learners — mastering logarithms starts with understanding their core equivalences!"]