Solve the following logarithmic equations: 1) log₂(x) + log₂(x-2) = 1 2) log₅(x+1) - log₅(x-1) = 1 3) log₃(x) + 2·log₃(x) = 6 4) log₄(x²-4) - log₄(x+2) = 1 5) log₂(x) + log₄(x) = 3 6) log(x²-1) = 2·log(x+1) 7) log₃(x) + log₃(x-2) + log₃(x-4) = log₃(15)

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Asked on August 10, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-08-10T06:30:15.118Z

Here are the solutions to the logarithm equations: 1) _2 x + _2 (x-2) = 1 Step 1: Use the logarithm property _b M + _b N = _b (MN). _2 (x(x-2)) = 1 Step 2: Convert the logarithmic equation to an exponential equation (M = b^N). x(x-2) = 2^1 x^2 - 2x = 2 x^2 - 2x - 2 = 0 Step 3: Solve the quadratic equation using the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a. x = -(-2) ± sqrt((-2)^2 - 4(1)(-2))2(1) x = 2 ± sqrt(4 + 8)2 x = 2 ± sqrt(12)2 x = 2 ± 2sqrt(3)2 x = 1 ± sqrt(3) Step 4: Check the domain restriction (x > 0 and x-2 > 0 x > 2). x = 1 + sqrt(3) ≈ 2.732 is valid. x = 1 - sqrt(3) ≈ -0.732 is not valid. The solution is x = 1 + sqrt(3). 2) _5 (x+1) - _5 (x-1) = 1 Step 1: Use the logarithm property _b M - _b N = _b ((M)/(N)). _5 ((x+1)/(x-1)) = 1 Step 2: Convert to an exponential equation. (x+1)/(x-1) = 5^1 x+1 = 5(x-1) x+1 = 5x - 5 Step 3: Solve for x. 6 = 4x x = (6)/(4) x = (3)/(2) Step 4: Check the domain restriction (x+1 > 0 x > -1 and x-1 > 0 x > 1, so x > 1). x = (3)/(2) = 1.5 is valid. The solution is x = (3)/(2). 3) _3 x + 2 · _3 x = 6 Step 1: Combine like terms. 3 · _3 x = 6 Step 2: Isolate _3 x. _3 x = (6)/(3) _3 x = 2 Step 3: Convert to an exponential equation. x = 3^2 x = 9 Step 4: Check the domain restriction (x > 0). x = 9 is valid. The solution is x = 9. 4) _4 (x^2-4) - _4 (x+2) = 1 Step 1: Use the logarithm property _b M - _b N = _b ((M)/(N)). _4 ((x^2-4)/(x+2)) = 1 Step 2: Factor the numerator x^2-4 = (x-2)(x+2). _4 (((x-2)(x+2))/(x+2)) = 1 Step 3: Simplify the expression inside the logarithm. The domain requires x > 2, so x+2 ≠ 0. _4 (x-2) = 1 Step 4: Convert to an exponential equation. x-2 = 4^1 x-2 = 4 x = 6 Step 5: Check the domain restriction (x^2-4 > 0 x < -2 or x > 2, and x+2 > 0 x > -2. Combined, x > 2). x = 6 is valid. The solution is x = 6. 5) _2 x + _4 x = 3 Step 1: Change the base of _4 x to base 2 using _b M = (_c M)/(_c b). _4 x = (_2 x)/(_2 4) = (_2 x)/(2) Step 2: Substitute this into the original equation. _2 x + (_2 x)/(2) = 3 Step 3: Combine the terms with _2 x. (1 + (1)/(2)) _2 x = 3 (3)/(2) _2 x = 3 Step 4: Isolate _2 x. _2 x = 3 · (2)/(3) _2 x = 2 Step 5: Convert to an exponential equation. x = 2^2 x = 4 Step 6: Check the domain restriction (x > 0). x = 4 is valid. The solution is x = 4. 6) (x^2-1) = 2 · (x+1) Step 1: Use the logarithm property k _b M = _b (M^k) on the right side. (x^2-1) = ((x+1)^2) Step 2: If _b M = _b N, then M = N. x^2-1 = (x+1)^2 Step 3: Expand the right side and solve for x. x^2-1 = x^2 + 2x + 1 -1 = 2x + 1 -2 = 2x x = -1 Step 4: Check the domain restriction (x^2-1 > 0 x < -1 or x > 1, and x+1 > 0 x > -1. Combined, x > 1). x = -1 is not greater than 1. Also, (x+1) would be (0), which is undefined. The equation has no solution. 7) _3 x + _3 (x-2) + _3 (x-4) = _3 (15) Step 1: Use the logarithm property _b M + _b N = _b (MN) on the left side. _3 (x(x-2)(x-4)) = _3 (15) Step 2: If _b M = _b N, then M = N. x(x-2)(x-4) = 15 Step 3: Expand the left side. x(x^2 - 4x - 2x + 8) = 15 x(x^2 - 6x + 8) = 15 x^3 - 6x^2 + 8x = 15 x^3 - 6x^2 + 8x - 15 = 0 Step 4: Find integer roots using the Rational Root Theorem. Test divisors of 15. Let P(x) = x^3 - 6x^2 + 8x - 15. P(5) = 5^3 - 6(5^2) + 8(5) - 15 = 125 - 150 + 40 - 15 = 0. So, x=5 is a root. Step 5: Perform polynomial division to factor (x-5) out. (x-5)(x^2 - x + 3) = 0 Step 6: Solve the quadratic factor x^2 - x + 3 = 0. The discriminant is = (-1)^2 - 4(1)(3) = 1 - 12 = -11. Since < 0, there are no other real roots. Step 7: Check the domain restriction (x > 0, x-2 > 0 x > 2, and x-4 > 0 x > 4. Combined, x > 4). x = 5 is valid. The solution is x = 5. Send me the next one 📸

Accepted Answer · 2026-08-10T06:30:15.118Z

Here are the solutions to the logarithm equations: 1) _2 x + _2 (x-2) = 1 Step 1: Use the logarithm property _b M + _b N = _b (MN). _2 (x(x-2)) = 1 Step 2: Convert the logarithmic equation to an exponential equation (M = b^N). x(x-2) = 2^1 x^2 - 2x = 2 x^2 - 2x - 2 = 0 Step 3: Solve the quadratic equation using the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a. x = -(-2) ± sqrt((-2)^2 - 4(1)(-2))2(1) x = 2 ± sqrt(4 + 8)2 x = 2 ± sqrt(12)2 x = 2 ± 2sqrt(3)2 x = 1 ± sqrt(3) Step 4: Check the domain restriction (x > 0 and x-2 > 0 x > 2). x = 1 + sqrt(3) ≈ 2.732 is valid. x = 1 - sqrt(3) ≈ -0.732 is not valid. The solution is x = 1 + sqrt(3). 2) _5 (x+1) - _5 (x-1) = 1 Step 1: Use the logarithm property _b M - _b N = _b ((M)/(N)). _5 ((x+1)/(x-1)) = 1 Step 2: Convert to an exponential equation. (x+1)/(x-1) = 5^1 x+1 = 5(x-1) x+1 = 5x - 5 Step 3: Solve for x. 6 = 4x x = (6)/(4) x = (3)/(2) Step 4: Check the domain restriction (x+1 > 0 x > -1 and x-1 > 0 x > 1, so x > 1). x = (3)/(2) = 1.5 is valid. The solution is x = (3)/(2). 3) _3 x + 2 · _3 x = 6 Step 1: Combine like terms. 3 · _3 x = 6 Step 2: Isolate _3 x. _3 x = (6)/(3) _3 x = 2 Step 3: Convert to an exponential equation. x = 3^2 x = 9 Step 4: Check the domain restriction (x > 0). x = 9 is valid. The solution is x = 9. 4) _4 (x^2-4) - _4 (x+2) = 1 Step 1: Use the logarithm property _b M - _b N = _b ((M)/(N)). _4 ((x^2-4)/(x+2)) = 1 Step 2: Factor the numerator x^2-4 = (x-2)(x+2). _4 (((x-2)(x+2))/(x+2)) = 1 Step 3: Simplify the expression inside the logarithm. The domain requires x > 2, so x+2 ≠ 0. _4 (x-2) = 1 Step 4: Convert to an exponential equation. x-2 = 4^1 x-2 = 4 x = 6 Step 5: Check the domain restriction (x^2-4 > 0 x < -2 or x > 2, and x+2 > 0 x > -2. Combined, x > 2). x = 6 is valid. The solution is x = 6. 5) _2 x + _4 x = 3 Step 1: Change the base of _4 x to base 2 using _b M = (_c M)/(_c b). _4 x = (_2 x)/(_2 4) = (_2 x)/(2) Step 2: Substitute this into the original equation. _2 x + (_2 x)/(2) = 3 Step 3: Combine the terms with _2 x. (1 + (1)/(2)) _2 x = 3 (3)/(2) _2 x = 3 Step 4: Isolate _2 x. _2 x = 3 · (2)/(3) _2 x = 2 Step 5: Convert to an exponential equation. x = 2^2 x = 4 Step 6: Check the domain restriction (x > 0). x = 4 is valid. The solution is x = 4. 6) (x^2-1) = 2 · (x+1) Step 1: Use the logarithm property k _b M = _b (M^k) on the right side. (x^2-1) = ((x+1)^2) Step 2: If _b M = _b N, then M = N. x^2-1 = (x+1)^2 Step 3: Expand the right side and solve for x. x^2-1 = x^2 + 2x + 1 -1 = 2x + 1 -2 = 2x x = -1 Step 4: Check the domain restriction (x^2-1 > 0 x < -1 or x > 1, and x+1 > 0 x > -1. Combined, x > 1). x = -1 is not greater than 1. Also, (x+1) would be (0), which is undefined. The equation has no solution. 7) _3 x + _3 (x-2) + _3 (x-4) = _3 (15) Step 1: Use the logarithm property _b M + _b N = _b (MN) on the left side. _3 (x(x-2)(x-4)) = _3 (15) Step 2: If _b M = _b N, then M = N. x(x-2)(x-4) = 15 Step 3: Expand the left side. x(x^2 - 4x - 2x + 8) = 15 x(x^2 - 6x + 8) = 15 x^3 - 6x^2 + 8x = 15 x^3 - 6x^2 + 8x - 15 = 0 Step 4: Find integer roots using the Rational Root Theorem. Test divisors of 15. Let P(x) = x^3 - 6x^2 + 8x - 15. P(5) = 5^3 - 6(5^2) + 8(5) - 15 = 125 - 150 + 40 - 15 = 0. So, x=5 is a root. Step 5: Perform polynomial division to factor (x-5) out. (x-5)(x^2 - x + 3) = 0 Step 6: Solve the quadratic factor x^2 - x + 3 = 0. The discriminant is = (-1)^2 - 4(1)(3) = 1 - 12 = -11. Since < 0, there are no other real roots. Step 7: Check the domain restriction (x > 0, x-2 > 0 x > 2, and x-4 > 0 x > 4. Combined, x > 4). x = 5 is valid. The solution is x = 5. Send me the next one 📸

Solve the following logarithmic equations: 1) log₂(x) + log₂(x-2) = 1 2) log₅(x+1) - log₅(x-1) = 1 3) log₃(x) + 2·log₃(x) = 6 4) log₄(x²-4) - log₄(x+2) = 1 5) log₂(x) + log₄(x) = 3 6) log(x²-1) = 2·log(x+1) 7) log₃(x) + log₃(x-2) + log₃(x-4) = log₃(15)

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Solve the following logarithmic equations: 1) log₂(x) + log₂(x-2) = 1 2) log₅(x+1) - log₅(x-1) = 1 3) log₃(x) + 2·log₃(x) = 6 4) log₄(x²-4) - log₄(x+2) = 1 5) log₂(x) + log₄(x) = 3 6) log(x²-1) = 2·log(x+1) 7) log₃(x) + log₃(x-2) + log₃(x-4) = log₃(15)

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