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.

Advanced30 min18 problems
Practice Problems

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.

Try it now

Which equation is the exponential form of log⁡4x=3\log_4 x = 3?

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}.

Interactive Visual

Function graph

051015-2024xy(8, 3)x = 0

Tap or hover over the graph to read off values.

y = log₂(x)y = 3Asymptote
  • Intersection point: (8, 3)

Table of values

x124816
log₂(x)01234
333333

Function graph

-6-4-20246810-4-20246xy(-2, 3)(4, 3)x = 0x = 2

Tap or hover over the graph to read off values.

Tap a name to show or hide its graph.

Asymptote
  • Intersection point: (-2, 3)
  • Intersection point: (4, 3)
  • The graphs lie on top of each other: infinitely many common points.

Interactive Sandbox

Interactive Grapher

-10-50510-10-50510xy

Tap or hover over the graph to read off values.

Try these examples:

y = 2x + 1

m=2, b=1

Expression Calculator

Try these:

History

No calculations yet

Practice Problems

18 problems
Problem 1 of 18
Easy
Which equation is the exponential form of log⁡4x=3\log_4 x = 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

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.
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.
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.
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.
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 $3^x = 20$.

Today's challengeOne problem, three tries

More in This Topic

Introduction to Logarithms

Advanced25 min

Definition of a Logarithm

Advanced25 min

Common Logarithms (Base 10)

Advanced25 min

Natural Logarithms (ln)

Advanced25 min