Solve logarithmic equations and check the domain

About this lesson

Algebra II › Logarithms

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Solve logarithmic equations and check the domain

Logarithms. log rules, change of base, logarithmic equations, half-life

Warm-up

Three quick questions from earlier lessons.

Discover

Solve log⁡2x=3\log_2 x = 3 by rewriting it as a power.

Then think: can ln⁡x\ln x ever be defined for x=−2x = -2? What does that mean for solutions you find?

Key idea

To solve a logarithmic equation: combine into one logarithm, rewrite as a power (or compare arguments), solve, then check the domain.

log⁡b(expression)=n  ⟺  expression=bn\log_b(\text{expression}) = n \iff \text{expression} = b^{n}, and every argument of a logarithm must be positive

Worked example

Solve ln⁡x+ln⁡(x−2)=ln⁡15\ln x + \ln(x - 2) = \ln 15.

ln⁡(x(x−2))=ln⁡15\ln\big(x(x - 2)\big) = \ln 15
Product rule.
x2−2x−15=0x^2 - 2x - 15 = 0
Equal logarithms have equal arguments.
(x−5)(x+3)=0(x - 5)(x + 3) = 0
Factor.
x=5x = 5
x = −3 makes ln x undefined: reject it.

Your turn, guided

Your turn: solve log⁡2(x−3)=4\log_2(x - 3) = 4. Fill in the missing line.

x−3=24=16x - 3 = 2^4 = 16
Rewrite as a power.
x=19x = 19
Add 3. Check: 19 − 3 = 16 > 0.

Practice ★

  1. Put the steps for solving ln⁡x+ln⁡(x−5)=ln⁡6\ln x + \ln(x - 5) = \ln 6 in order.
    • Combine with the product rule: ln⁡(x(x−5))=ln⁡6\ln\big(x(x - 5)\big) = \ln 6
    • Equal logarithms have equal arguments: x(x−5)=6x(x - 5) = 6
    • Solve the quadratic equation
    • Keep only the values where x>5x > 5
  2. Put the steps for solving ln⁡x+ln⁡(x−4)=ln⁡21\ln x + \ln(x - 4) = \ln 21 in order.
    • Combine with the product rule: ln⁡(x(x−4))=ln⁡21\ln\big(x(x - 4)\big) = \ln 21
    • Equal logarithms have equal arguments: x(x−4)=21x(x - 4) = 21
    • Solve the quadratic equation
    • Keep only the values where x>4x > 4
  1. Solve log⁡5(x−8)=4\log_{ 5 }(x - 8) = 4.
  2. Solve log⁡5(x−9)=3\log_{ 5 }(x - 9) = 3.
  3. Solve log⁡3(x−8)=4\log_{ 3 }(x - 8) = 4.
  4. Solve log⁡2(x−2)=4\log_{ 2 }(x - 2) = 4.
  5. Solve log⁡2(x−6)=4\log_{ 2 }(x - 6) = 4.

Watch out!

Two learners solved ln⁡x+ln⁡(x−3)=ln⁡4\ln x + \ln(x - 3) = \ln 4. Compare their work.

Solution A

x2−3x−4=0x^2 - 3x - 4 = 0
Product rule, then compare the arguments.
(x−4)(x+1)=0(x - 4)(x + 1) = 0
Factor.
x=4x = 4 or x=−1x = -1
Both values solve the quadratic.

Solution B

x2−3x−4=0x^2 - 3x - 4 = 0
Product rule, then compare the arguments.
(x−4)(x+1)=0(x - 4)(x + 1) = 0
Factor.
x=4x = 4
ln(−1) is not defined, so x = −1 is rejected.

  1. Solution A is correct
  2. Solution B is correct
  3. Both are correct

Practice ★★

  1. Solve ln⁡x+ln⁡(x−1)=ln⁡42\ln x + \ln(x - 1) = \ln 42. Give every solution.
  2. Solve ln⁡x+ln⁡(x−2)=ln⁡24\ln x + \ln(x - 2) = \ln 24. Give every solution.
  3. Solve ln⁡x+ln⁡(x−3)=ln⁡18\ln x + \ln(x - 3) = \ln 18. Give every solution.
  4. Solve ln⁡x+ln⁡(x−4)=ln⁡45\ln x + \ln(x - 4) = \ln 45. Give every solution.
  5. Solve ln⁡x+ln⁡(x−1)=ln⁡72\ln x + \ln(x - 1) = \ln 72. Give every solution.

Exit ticket

Three questions to check you have got it.

  1. Solve ln⁡x+ln⁡(x−2)=ln⁡3\ln x + \ln(x - 2) = \ln 3.
  2. Solve ln⁡x+ln⁡(x−2)=ln⁡48\ln x + \ln(x - 2) = \ln 48.
  3. Solve ln⁡x+ln⁡(x−1)=ln⁡56\ln x + \ln(x - 1) = \ln 56.

Summary card

Solving logarithmic equations

  1. Combine into one logarithm (product, quotient, power rules).
  2. Rewrite as a power, or compare arguments.
  3. Solve, then check that every argument is positive.

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