Logarithmic Equation Resolution

Logarithmic Properties and Equations

Key Concepts

  • Logarithmic Identity:
    • Logarithms allow the combination of logarithmic terms using properties of logarithms.
  • Equality:
    • Establishing equality through the use of logarithmic identities is critical in solving logarithmic equations.

Detailed Breakdown of the Equation

  • The given equation is:
    log(3x+2)+log(x1)=1\log(3x + 2) + \log(x - 1) = 1
Applying Logarithmic Properties
  • Sum of Logarithms:

    • The sum of two logarithms can be combined into a single logarithm:
      log(a)+log(b)=log(ab)\log(a) + \log(b) = \log(ab)
    • Applying this property:
      log(3x+2)+log(x1)=log((3x+2)(x1))\log(3x + 2) + \log(x - 1) = \log((3x + 2)(x - 1))
  • Therefore, we can rewrite the equation as:
    log((3x+2)(x1))=1\log((3x + 2)(x - 1)) = 1

Converting from Logarithmic to Exponential Form
  • Exponential Form:
    • To solve for x, we convert the logarithmic equation into its exponential form:
      a=logb(c)    ba=ca = \log_b(c) \implies b^a = c
    • Here, base 10 is implicit. Hence:
      101=(3x+2)(x1)10^1 = (3x + 2)(x - 1)
    • Which simplifies to:
      10=(3x+2)(x1)10 = (3x + 2)(x - 1)

Final Steps to Solve for x

  • Expand and Rearrange:

    • Expanding the right side:
      10=3x23x+2x210 = 3x^2 - 3x + 2x - 2
    • Simplifies to:
      10=3x2x210 = 3x^2 - x - 2
  • Setting Up the Quadratic Equation:

    • Rearranging gives:
      3x2x12=03x^2 - x - 12 = 0
  • Solving the Quadratic:

    • Apply the quadratic formula:
      x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
    • Where:
    • a = 3, b = -1, c = -12
  • Calculating:
    x=(1)±(1)24(3)(12)2(3)x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-12)}}{2(3)}

    • Leads to:
      x=1±1+1446x = \frac{1 \pm \sqrt{1 + 144}}{6}
    • Further simplifying gives:
      x=1±136x = \frac{1 \pm 13}{6}
Possible Solutions
  • The two potential solutions derived are:
    1. x=146=73x = \frac{14}{6} = \frac{7}{3}
    2. x=126=2x = \frac{-12}{6} = -2
Verifying Solutions
  • Since logarithms are not defined for negative inputs or zero, check which solution is valid:
    • For x=73x = \frac{7}{3}:
    • Both 3x+23x + 2 and x1x - 1 are positive.
    • For x=2x = -2:
    • This is invalid because x - 1 = -3 < 0.

Conclusion

  • Valid Solution:
    • The only acceptable solution to the equation is:
      x=73x = \frac{7}{3}