Solve Logarithmic Equations - Detailed Solutions

Solve logarithmic equations including some challenging questions. Detailed solutions are presented. The logarithmic equations in examples 4, 5, 6 and 7 involve logarithms with different bases and are therefore challenging.

Example 1: Solve the logarithmic equation

log2(x - 1) = 5.
Solution to example 1
  • Rewrite the logarithm as an exponential using the definition.

    x - 1 = 25

  • Solve the above equation for x.

    x = 33

  • check:

    Left Side of equation
    log2(x - 1) = log2(33 - 1) = log2(25) = 5

    Right Side of equation = 5

  • conclusion: The solution to the above equation is x = 33

Example 2: Solve the logarithmic equation

log5(x - 2) + log5(x + 2) = 1.
Solution to example 2
  • Use the product rule to the expression in the right side.

    log5(x - 2)(x + 2) = 1

  • Rewrite the logarithm as an exponential (definition).

    (x - 2)(x + 2) = 51

  • Which can be simplified as.

    x2 = 9

  • Solve for x.

    x = 3 and x = -3

  • check: 1st solution x = 3

    Left Side of equation: log5(3 - 2) + log5(3 + 2) = log51 + log5(3 + 2) = log55 = 1

    Right Side of Equation

    2nd solution x = -3
    Left Side of equation: log5(-3 - 2) + log5(-3 + 2) = log5(-5) + log5(-1)

    log5(-5) and log5(-1) are both undefined and therefore x = -3 is not a solution.

    conclusion: The solutions to the given equation is x = 3


Example 3: Solve the logarithmic equation

log3(x - 2) + log3(x - 4) = log3(2x^2 + 139) - 1.
Solution to example 3
  • We first replace 1 in the equation by log3(3) and rewrite the equation as follows.

    log3(x - 2) + log3(x - 4) = log3(2x^2 + 139) - log3(3)

  • We now use the product and quotient rules of the logarithm to rewrite the equation as follows.

    log3[ (x - 2)(x - 4) ] = log3[ (2x^2 + 139) / 3 ]

  • Which gives the algebraic equation

    (x - 2)(x - 4) = (2x^2 + 139) / 3

  • Mutliply all terms by 3 and simplify

    3(x - 2)(x - 4) = (2x^2 + 139)

  • Solve the above quadratic equation to obtain

    x = -5 and x = 23


  • check:

    1) x = - 5 cannot be a solution to the given equation as it would make the argument of the logarithmic functions on the right negative.

    2) x = 23

    Right Side of equation:

    log3(23 - 2) + log3(23 - 4) = log3(21*19) = log3(399)

    Left Side of equation:

    log3(2(23)^2 + 139) - 1 = log3(1197) - log3(3) = log3(1197 / 3) = log3(399)

  • conclusion: The solution to the above equation is x = 23

Example 4: Solve the logarithmic equation

log4(x + 1) + log16(x + 1) = log4(8).
Solution to example 4
  • We first note that 2 logarithms in the given equation have base 4 and one has base 16. We first use the change of base formula to write that

    log16(x + 1) = log4(x + 1) / log4(16) = log4(x + 1) / 2 = log4(x + 1)1/2

  • We now write the given equation as follows.

    log4(x + 1) + log4(x + 1)1/2 = log4(8)

  • We use the product rule to write.

    log4(x + 1)(x + 1)1/2 = log4(8)

  • Which gives

    (x + 1)(x + 1)1/2 = (8)

  • which can be written as

    (x + 1)3/2 = (8)

  • Solve for x to obtain.

    x = 3

  • check:

    Left Side of equation:
    log4(3 + 1) + log16(3 + 1) = 1 + 1/2 = 3/2

    Right Side of equation:
    log4(8) = log4(43/2) = 3/2
  • conclusion: The solution to the above equation is x = 3

Example 5: Solve the logarithmic equation

log2(x - 4) + logsqrt(2)(x3 - 2) + log0.5(x - 4) = 20.
Solution to example 5
  • We first use the change of base formula to write.

    logsqrt(2)(x3 - 2) = log2(x3 - 2) / log2(sqrt(2)) = 2log2(x3 - 2)

  • We also use the change of base formula to write.

    log0.5(x - 4) = -log2(x - 4)

  • Substitute into the equation and simplify the given equation.

    2 log2(x3 - 2) = 20

  • rewrite as.

    log2(x3 - 2) = 10

  • which gives

    x3 - 2 = 210

  • Solve the above equation for x.

    x = cube_root (1026)

Example 6: Solve the logarithmic equation

ln(x + 6) + log(x + 6) = 4.
Solution to example 6
  • Use the change of base formula to rewrite log(x + 6) as

    log(x + 6) = ln(x + 6) / ln(10)

  • and substitute in the given equation

    ln(x + 6) + ln(x + 6) / ln(10) = 4

  • solve for ln(x + 6)

    ln(x + 6) = 4 ln(10) / (1 + ln(10))

  • solve the above for x

    x = e4 ln(10) / (1 + ln(10)) - 6

Example 7: Solve the logarithmic equation

log5(ln(x + 3) - 1) + log1/5(ln(x + 3) - 1) = 0.
Solution to example 7
  • The change of base formula is used to write

    log1/5(ln(x + 3) = -log5(ln(x + 3)

  • Substitute in the given equation

    log5(ln(x + 3) - 1) - log5(ln(x + 3) - 1) = 0

  • The left hand term is equal to 0 for x + 3 > 0 and ln(x + 3) - 1 > 0.

    x + 3 > 0 gives x > -3

  • ln(x + 3) - 1 > 0 gives.

    ln(x + 3) > 1

  • or

    x + 3 > e

  • or

    x > e - 3

  • conclusion: The solution set to the above equation is given by the interval (e - 3 , + infinity). It is an identity.

More topics to explore and Tests:
Experiment and Explore Mathematics: Tutorials and Problems

Solve Exponential and Logarithmic Equations

Solve Exponential and Logarithmic Equations - Tutorial

Logarithmic Functions (with applet)


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Updated: 2 April 2013

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