Find the Exact Solution in Therms of Logarithms

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

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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

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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

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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(4^3/2) = 3/2

• conclusion: The solution to the above equation is x = 3

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Example 5:

Solve the logarithmic equation

log₂(x - 4) + log_sqrt(2)(x³ - 2) + log_0.5(x - 4) = 20. Solution to example 5

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

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

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

  log_0.5(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)

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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 = e^{4 ln(10) / (1 + ln(10))} - 6

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Example 7:

Solve the logarithmic equation

log₅(ln(x + 3) - 1) + log_1/5(ln(x + 3) - 1) = 0. Solution to example 7

• The change of base formula is used to write

  log_1/5(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.

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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)

Find the Exact Solution in Therms of Logarithms

Source: https://www.analyzemath.com/logarithmic_equations/solve.html

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