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Teacher Guide: Solving Logarithmic Equations

Solve equations with the unknown inside a logarithm, reject extraneous solutions, and use logarithms to solve exponential equations like 3^x = 20.

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10 questions on Logarithms. Students join with a name, you see everyone's score.

For Teachers

Learning Objectives
  • Solve logarithmic equations by rewriting them in exponential form
  • Solve equations with logarithms on both sides using the one-to-one property
  • Condense sums and differences of logarithms before solving, and identify and reject extraneous solutions
  • Solve exponential equations with different bases by taking logarithms, giving exact and rounded answers
Prerequisites
  • • Definition of a Logarithm (converting between logarithmic and exponential form)
  • • Product, Quotient and Power Rule for Logarithms
  • • Change of Base Formula
  • • Solving Exponential Equations (same-base method)
  • • Solving linear and quadratic equations by factoring
Discussion Starters
  • 1. Why may you remove the logarithms in log⁡5A=log⁡5B\log_5 A = \log_5 B, but not in log⁡5A+log⁡5B=2\log_5 A + \log_5 B = 2?
  • 2. The equation log⁡2x+log⁡2(x−2)=3\log_2 x + \log_2(x - 2) = 3 gives the candidates 44 and −2-2. Where does the second one come from?
  • 3. Can log⁡3x=−2\log_3 x = -2 have a solution even though the right side is negative? What about log⁡3x=0\log_3 x = 0?
  • 4. Before using the calculator, how can you tell that the solution of 3x=203^x = 20 lies between 22 and 33?
Common Misconceptions

Every number found while solving is a solution

Remediation: Make the domain check a fixed last step. Show both graphs of log⁡2x+log⁡2(x−2)\log_2 x + \log_2(x - 2) and log⁡2(x(x−2))\log_2\big(x(x - 2)\big): they agree for x>2x > 2, but only the condensed one exists for x<0x < 0.

A negative value of xx is always extraneous

Remediation: Use log⁡2(x+5)=1\log_2(x + 5) = 1, where x=−3x = -3 is a valid solution. Emphasize that the arguments must be positive, not xx itself.

log⁡20log⁡3\frac{\log 20}{\log 3} equals log⁡203\log \frac{20}{3}

Remediation: Let students evaluate both on the calculator: 2.732.73 versus about 0.820.82. The quotient rule is about the logarithm of a quotient, not a quotient of logarithms.

Differentiation Ideas

For Struggling Students:

  • • Start with direct equations like log⁡2x=5\log_2 x = 5 and practice only the conversion to exponential form
  • • Give a fixed checklist: condense, rewrite, solve, check the domain
  • • Provide a table of powers of 2, 3, 5 and 10 to recognize values like 81=3481 = 3^4

For On-Level Students:

  • • Solve equations with logarithms on both sides and with an expression inside the logarithm
  • • Solve equations that condense to a quadratic and reject the extraneous candidate
  • • Solve exponential equations like 5x=405^x = 40 and round the answer to two decimal places

For Advanced Students:

  • • Solve equations with logarithms of different bases by first applying the change of base formula
  • • Solve exponential equations with unknowns on both sides, such as 2x+1=3x2^{x+1} = 3^x, by taking logarithms
  • • Construct an equation where both candidates are extraneous and explain why it has no solution
Standards Alignment
  • HSF-LE.A.4 (CCSS.MATH.CONTENT.HSF.LE.A.4)

    For exponential models, express as a logarithm the solution to ab^(ct) = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology

  • HSF-BF.B.5 (CCSS.MATH.CONTENT.HSF.BF.B.5)

    Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents

  • HSA-REI.A.2 (CCSS.MATH.CONTENT.HSA.REI.A.2)

    Solve simple equations in one variable and give examples showing how extraneous solutions may arise

  • HSA-REI.D.11 (CCSS.MATH.CONTENT.HSA.REI.D.11)

    Explain why the x-coordinates of the points where the graphs of y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x)

Lesson Resources
  • visualTwo Graphs, One Equation

    Students compare the graphs of the original and the condensed equation and see where the extraneous solution comes from

  • activityAccept or Reject

    Cards with an equation and a candidate: students substitute and decide whether every argument stays positive

  • worksheetLog Equation Practice

    Equations from exponential form to condensing with extraneous solutions, with a domain-check column

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

A logarithmic equation is an equation in which the unknown appears inside a logarithm, for example log⁡2x=3\log_2 x = 3 or log⁡3(2x+1)=2\log_3(2x + 1) = 2.
Two tools solve almost every logarithmic equation:
1. Rewrite in exponential form. A logarithm is an exponent, so
log⁡bA=c  ⟺  A=bc\log_b A = c \iff A = b^c
For example, log⁡2x=3\log_2 x = 3 means x=23=8x = 2^3 = 8. In the graph, the line y=3y = 3 meets the curve y=log⁡2xy = \log_2 x exactly at x=8x = 8.
2. Use the one-to-one property. If two logarithms with the same base are equal, their arguments are equal:
log⁡bM=log⁡bN  ⟹  M=N\log_b M = \log_b N \implies M = N
The method:
  1. 1.Condense each side into a single logarithm with the product, quotient and power rules.
  2. 2.Rewrite in exponential form, or set the arguments equal.
  3. 3.Solve the resulting equation (usually linear or quadratic).
  4. 4.Check the domain: every argument in the ORIGINAL equation must be positive. A candidate that makes an argument zero or negative is an extraneous solution and must be rejected.
Exponential equations with different bases. When the two sides of an exponential equation cannot be written with the same base, take the logarithm of both sides and use the power rule:
3x=20  ⟹  xlog⁡3=log⁡20  ⟹  x=log⁡20log⁡3≈2.733^x = 20 \implies x \log 3 = \log 20 \implies x = \frac{\log 20}{\log 3} \approx 2.73
Here log⁡\log is the common logarithm (base 10). The natural logarithm ln⁡\ln gives the same result.

Worked Examples

Solve log⁡2x=3\log_2 x = 3.
1

Identify the parts

Base 22, argument xx, value of the logarithm 33 → log⁡2x=3\log_2 x = 3
2

Rewrite in exponential form

log⁡bA=c  ⟺  A=bc\log_b A = c \iff A = b^c → x=23x = 2^3
3

Calculate the power

23=2⋅2⋅22^3 = 2 \cdot 2 \cdot 2 → x=8x = 8
4

Check

log⁡28=3\log_2 8 = 3 because 23=82^3 = 8 ✓, and the argument 88 is positive → The solution is valid

Common Mistakes

Keeping both candidates: log⁡2x+log⁡2(x−2)=3  ⟹  x=4\log_2 x + \log_2(x - 2) = 3 \implies x = 4 or x=−2x = -2

Why it's wrong: After condensing to log⁡2(x(x−2))\log_2\big(x(x - 2)\big), negative values of xx seem allowed, because the product of two negative numbers is positive. But the original equation contains log⁡2x\log_2 x, which is not defined for x=−2x = -2.

Correct: Check every candidate in the ORIGINAL equation. Only x=4x = 4 works; x=−2x = -2 is extraneous.

Dropping the logarithms: log⁡2x+log⁡2(x−2)=3  ⟹  x+(x−2)=3\log_2 x + \log_2(x - 2) = 3 \implies x + (x - 2) = 3

Why it's wrong: You may remove logarithms only when each side is a single logarithm with the same base (one-to-one property). Here the left side is a sum of logarithms and the right side is a number.

Correct: Condense first, log⁡2(x(x−2))=3\log_2\big(x(x - 2)\big) = 3, then rewrite in exponential form: x(x−2)=23=8x(x - 2) = 2^3 = 8.

Swapping base and exponent: log⁡3(2x+1)=2  ⟹  2x+1=23\log_3(2x + 1) = 2 \implies 2x + 1 = 2^3

Why it's wrong: In log⁡bA=c\log_b A = c the base bb is raised to the power cc. The base of the logarithm stays the base of the power.

Correct: 2x+1=32=92x + 1 = 3^2 = 9, so x=4x = 4. Check: log⁡39=2\log_3 9 = 2 ✓

Dividing instead of taking logarithms: 3x=20  ⟹  x=203≈6.673^x = 20 \implies x = \frac{20}{3} \approx 6.67

Why it's wrong: 3x3^x is a power, not the product 3⋅x3 \cdot x. A quick check shows the error: 36.673^{6.67} is about 15001500, far more than 2020.

Correct: x=log⁡20log⁡3≈2.73x = \frac{\log 20}{\log 3} \approx 2.73. Note that log⁡20log⁡3\frac{\log 20}{\log 3} is NOT the same as log⁡203\log \frac{20}{3}.

Why It Matters

Logarithmic equations answer "how long?" and "how strong?" questions whenever something grows or shrinks by a constant factor:
  • Finance: How many years until savings at 8% interest double? Solve 1.08t=21.08^t = 2.
  • Biology: When does a bacteria culture reach a target size that is not a power of 2?
  • Sound: How intense is a 90 dB lawn mower compared with the quietest sound you can hear?
  • Chemistry and geology: pH values and earthquake magnitudes are logarithms, so finding the concentration or energy behind them means solving a logarithmic equation.
In the lesson Solving Exponential Equations you could only say that the solution of 2x=202^x = 20 lies between 44 and 55. With logarithms you get the exact value x=log⁡20log⁡2≈4.32x = \frac{\log 20}{\log 2} \approx 4.32. And checking the domain is a habit you will need again for every equation with roots, fractions or logarithms.

Real World Applications

Doubling Your Savings

Money that grows by a fixed percentage each year follows $A = P \cdot (1 + r)^t$. Solving for the exponent $t$ with logarithms tells you when a savings goal is reached.

Example:

At 6% interest a year, money doubles when 1.06t=21.06^t = 2, so t=ln⁡2ln⁡1.06≈11.9t = \frac{\ln 2}{\ln 1.06} \approx 11.9 years.

1Try It Yourself

You put 1000 dollars into a savings account that pays 8% interest a year, so after tt years you have 1000⋅1.08t1000 \cdot 1.08^t dollars.

After how many years has the money doubled to 2000 dollars?

Step 1: Write the mathematical expression

Solve 1000⋅1.08t=20001000 \cdot 1.08^t = 2000 and write the exact value of tt as a quotient of two logarithms, in the form ln(a)/ln(b) (do not round yet):

Loudness in Decibels

The sound level in decibels is $L = 10 \log r$, where $r = \frac{I}{I_0}$ compares the intensity of a sound with the quietest sound people can hear.

Example:

Normal conversation is about 60 dB: 10log⁡r=6010 \log r = 60 gives log⁡r=6\log r = 6, so r=106r = 10^6, one million times the threshold of hearing.

2Try It Yourself

A lawn mower measures 90 dB. Its intensity ratio rr satisfies 10log⁡r=9010 \log r = 90.

What is the intensity ratio rr?

Step 1: Write the mathematical expression

Write rr as a power of 10:

Growing Bacteria

A colony that doubles every hour follows $N = N_0 \cdot 2^t$. When the target is not a power of 2, logarithms give the exact time.

Example:

A colony of 200 cells reaches 1000 cells when 2t=52^t = 5, so t=ln⁡5ln⁡2≈2.32t = \frac{\ln 5}{\ln 2} \approx 2.32 hours.

3Try It Yourself

A culture starts with 500 bacteria and doubles every hour.

After how many hours are there 6000 bacteria?

Step 1: Write the mathematical expression

Solve 500⋅2t=6000500 \cdot 2^t = 6000 and write the exact value of tt as a quotient of two logarithms, in the form ln(a)/ln(b) (do not round yet):

Key Takeaways

  • 1A logarithmic equation has the unknown inside a logarithm, like log⁡2x=3\log_2 x = 3
  • 2Rewrite in exponential form: log⁡bA=c  ⟺  A=bc\log_b A = c \iff A = b^c
  • 3One-to-one property: log⁡bM=log⁡bN\log_b M = \log_b N gives M=NM = N
  • 4Condense sums and differences of logarithms into a single logarithm before you solve
  • 5Check every candidate in the ORIGINAL equation: all arguments must be positive, otherwise the candidate is an extraneous solution
  • 6For exponential equations with different bases, take logarithms: 3x=20  ⟹  x=log⁡20log⁡3≈2.733^x = 20 \implies x = \frac{\log 20}{\log 3} \approx 2.73

Frequently Asked Questions

Why do extraneous solutions appear at all?

Condensing changes the domain. log⁡2x+log⁡2(x−2)\log_2 x + \log_2(x - 2) needs x>2x > 2, but log⁡2(x(x−2))\log_2\big(x(x - 2)\big) only needs x(x−2)>0x(x - 2) > 0, which is also true for x<0x < 0. The condensed equation can therefore have solutions the original one does not have. That is why you always check in the original equation.

Is a negative solution always extraneous?

No. What matters is the sign of each argument, not the sign of xx. In log⁡2(x+5)=1\log_2(x + 5) = 1 we get x+5=2x + 5 = 2, so x=−3x = -3, and the argument −3+5=2-3 + 5 = 2 is positive. The solution x=−3x = -3 is valid.

Should I use log or ln to solve 3x=203^x = 20?

Either one. log⁡20log⁡3\frac{\log 20}{\log 3} and ln⁡20ln⁡3\frac{\ln 20}{\ln 3} give the same number, about 2.732.73. This is exactly the change of base formula: both are equal to log⁡320\log_3 20.

Can a logarithmic equation have no solution?

Yes. For log⁡2(x−4)=log⁡2(2x−1)\log_2(x - 4) = \log_2(2x - 1) the arguments give x−4=2x−1x - 4 = 2x - 1, so x=−3x = -3. But then both arguments are −7-7, which is not allowed. The only candidate is rejected, so the equation has no solution.

Glossary

Logarithmic equation
An equation in which the unknown appears inside a logarithm, such as log⁡3(2x+1)=2\log_3(2x + 1) = 2
Exponential form
The power form of a logarithm: log⁡bA=c\log_b A = c is written as A=bcA = b^c
One-to-one property
If log⁡bM=log⁡bN\log_b M = \log_b N, then M=NM = N (for M,N>0M, N > 0)
Candidate
A value found while solving that still has to be checked in the original equation
Extraneous solution
A candidate that makes an argument of the original equation zero or negative and is therefore rejected
Domain of a logarithm
The values for which log⁡bA\log_b A is defined: the argument must be positive, A>0A > 0

Formula Card

Exponential Form

log⁡bA=c  ⟺  A=bc\log_b A = c \iff A = b^c

Valid for $b > 0$, $b \neq 1$. Use it when one side is a single logarithm and the other side is a number.

One-to-One Property

log⁡bM=log⁡bN  ⟹  M=N\log_b M = \log_b N \implies M = N

Use it when both sides are single logarithms with the same base.

Domain Check

A>0A > 0

Every argument $A$ of a logarithm in the original equation must be positive. Otherwise the candidate is extraneous.

Logarithm of Both Sides

bx=c  ⟹  x=log⁡clog⁡b=ln⁡cln⁡bb^x = c \implies x = \frac{\log c}{\log b} = \frac{\ln c}{\ln b}

Solves exponential equations whose sides cannot be written with the same base, for example $3^x = 20$.

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