Grade by Grade
A. Exponential and Logarithmic Functions · Exponential and Logarithmic Functions

Solving Logarithmic Equations

A3.3 A3.1About 75 minutes
Learning goals
  • I can solve a logarithmic equation by rewriting it in exponential form.
  • I can solve an equation with one logarithm on each side by setting the arguments equal.
  • I can condense several logarithms into one before solving.
  • I can identify and reject extraneous solutions that make an argument zero or negative.

Words to know

logarithmic equation
An equation in which the variable appears inside a logarithm, such as log x + log (x − 3) = 1.
extraneous solution
A value produced by the algebra that does not satisfy the original equation. In log equations it usually makes an argument zero or negative.
equate the arguments
If log_b M = log_b N (M, N > 0), then M = N, because the log function is one-to-one.
admissible
Allowed: a solution is admissible if every logarithm in the original equation is defined for it.

Equations with the unknown inside a log

In lesson 3 you solved simple logarithmic equations such as log₃ x = 4 by switching to exponential form. Real equations often have two or more logarithms, like log x + log (x − 3) = 1. The plan is to condense them into one log using the laws, then convert to exponential form, solve and check.

The checking step is not optional. The laws of logarithms change the domain (as you saw in lesson 10), so the algebra can produce answers that do not work in the original equation.

1. Condenseinto one logon each side2. Convertto exponentialform (or equatearguments)3. Solvethe linear orquadratic4. Checkevery argumentmust be > 0
Four steps for logarithmic equations. The last step, checking that every argument in the original equation is positive, catches extraneous solutions.
Worked example

Example 1: One log, convert to exponential form

Solve log₃ (x + 1) = 2 and log (4x) = 3.

  1. log₃ (x + 1) = 2 means x + 1 = 3² = 9, so x = 8. Check: log₃ 9 = 2 ✓
  2. log (4x) = 3 means 4x = 10³ = 1000, so x = 250. Check: log 1000 = 3 ✓

x = 8 and x = 250

Worked example

Example 2: One log on each side, equate arguments

Solve log (2x + 1) = log (x + 4).

  1. Both sides are common logs, and log is one-to-one, so 2x + 1 = x + 4.
  2. x = 3.
  3. Check: 2(3) + 1 = 7 > 0 and 3 + 4 = 7 > 0, so both logs are defined ✓

x = 3

Worked example

Example 3: Condense, then solve a quadratic

Solve log x + log (x − 3) = 1.

  1. Product law: log (x(x − 3)) = 1.
  2. Exponential form: x(x − 3) = 10¹ = 10.
  3. x² − 3x − 10 = 0, so (x − 5)(x + 2) = 0, giving x = 5 or x = −2.
  4. Check x = 5: log 5 + log 2 = log 10 = 1 ✓
  5. Check x = −2: log (−2) is undefined. Reject x = −2; it is extraneous.

x = 5 only.

xy−6−4−2246810−2−112(5, 1)(−2, 1)y = 1y = log x + log (x − 3)y = log (x(x − 3))x = 3
The solid curve y = log x + log (x − 3) exists only for x > 3 and meets y = 1 once, at (5, 1). The dotted curve y = log (x(x − 3)) (the condensed form) lies exactly under the solid curve on the right, but it has an extra left branch that also meets y = 1, at (−2, 1). That is where the extraneous solution comes from.
Worked example

Example 4: A quotient

Solve log₂ (x + 2) − log₂ (x − 1) = 2.

  1. Quotient law: log₂ ((x + 2)/(x − 1)) = 2.
  2. Exponential form: (x + 2)/(x − 1) = 2² = 4.
  3. x + 2 = 4(x − 1) = 4x − 4, so 6 = 3x and x = 2.
  4. Check: log₂ 4 − log₂ 1 = 2 − 0 = 2 ✓

x = 2

Worked example

Example 5: A power on one side

Solve 2 log x = log (5x − 6).

  1. Power law: log (x²) = log (5x − 6). (This step is safe here because the original 2 log x already requires x > 0.)
  2. Equate arguments: x² = 5x − 6, so x² − 5x + 6 = 0 and (x − 2)(x − 3) = 0.
  3. x = 2: log 2 and log (10 − 6) = log 4 are defined, and 2 log 2 = log 4 ✓
  4. x = 3: 2 log 3 = log 9 and log (15 − 6) = log 9 ✓

x = 2 or x = 3. Not every quadratic produces an extraneous root; you have to check each one.

Worked example

Example 6: A quadratic in log x

Solve (log x)² − 3 log x + 2 = 0.

  1. Let u = log x: u² − 3u + 2 = 0, so (u − 1)(u − 2) = 0.
  2. log x = 1 gives x = 10; log x = 2 gives x = 100.
  3. Both are positive, so both are admissible.

x = 10 or x = 100

Where to check

Always substitute into the original equation, not the condensed one. In Example 3, x = −2 satisfies log (x(x − 3)) = 1 perfectly, because (−2)(−5) = 10. Only the original equation shows it is extraneous.

Just for fun

Riddle: Why was x = −2 not allowed into the logarithm party?

It couldn't show a positive argument.

Hands-on activity

Keep or reject?

You need: 12 index cards, a pencil and a scientific calculator.

  1. Copy each equation onto its own card: log₂ x + log₂ (x − 2) = 3; log (x + 6) − log (x − 3) = 1; log₃ (x² − 8x) = 2; log x + log (x + 21) = 2.
  2. Solve each one on scrap paper. You will get two candidate answers for most of them.
  3. Write each candidate on its own card. Test it in the original equation with your calculator.
  4. Sort the candidate cards into a KEEP pile and a REJECT pile, and write the reason on each rejected card.

What to look for: (1) x² − 2x = 8 gives x = 4 (keep) or −2 (reject: log₂ (−2) undefined). (2) (x + 6)/(x − 3) = 10 gives x = 4 (keep). (3) x² − 8x = 9 gives x = 9 or x = −1; both are kept, because the only argument is x² − 8x, which equals 9 > 0 for both. (4) x² + 21x = 100 gives x = 4 (keep) or −25 (reject). Rejected answers make an argument in the original equation zero or negative.

Your turn

1

Solve log₅ (2x − 1) = 2.

2

Solve log (3x − 2) = log (x + 6).

3

Solve log₂ x + log₂ (x + 2) = 3.

4

Solve log₃ (x + 7) − log₃ (x − 1) = 2.

5

Solve log (x + 1) + log (x − 2) = 1.

6

Solve 2 log₃ x = log₃ (x + 6).

7

Solve (log x)² = 4.

More practice: warm-up

1

Solve log₂ x = 5.

2

Solve log x = −1.

3

Solve log₄ (x − 1) = 1.

4

Solve log₇ (2x) = log₇ 18.

5

Solve log₃ x + log₃ 2 = log₃ 10.

More practice: on your own

1

Solve log₆ x + log₆ (x + 5) = 2.

2

Solve log (x + 4) − log x = log 3.

3

Solve log₂ (x² − 4x) = 5.

4

Solve 2 log (x − 1) = log (x + 5).

5

Solve log₂ (x + 1) = 3 − log₂ (x − 1).

Challenge

1

Solve log₃ x + log₉ x = 6. (Hint: write log₉ x in base 3.)

2

Solve log (x + 2) + log (x − 1) = log (3x + 6) − log 2.

3

Solve log₂ (log₃ x) = 1.

Check your understanding

Quiz

1.

Solve log₂ (x − 3) = 4.

2.

Solve log (5x) = log (2x + 9).

3.

Solving log x + log (x − 9) = 1 gives x = 10 or x = −1. Which is correct?

4.

Which equation results from condensing log₃ (x + 5) − log₃ x = 2 and converting to exponential form?

5.

Why must you check solutions of log equations in the original equation?

6.

Solve (log x)² − log x − 2 = 0.

For parents and teachers

Big idea (A3.3): to solve a logarithmic equation, condense to one log per side, convert to exponential form (or equate arguments), solve, and then check every candidate in the original equation.

How to help: After your teen solves, point at each log in the original equation and ask "Is its argument positive for this answer?" That habit catches every extraneous root.

Common difficulties:

  • Skipping the check, or checking in the condensed equation instead of the original.
  • Rejecting a negative x automatically. A negative x is fine if every argument is still positive (as in log₂ (x² − 4x) = 5, where x = −4 works).
  • Writing log x + log (x − 3) = 1 as x + (x − 3) = 10.
  • Forgetting that log x = −1 gives x = 0.1, a perfectly good positive answer.