What is a logarithmic equation?

A logarithmic equation is an equation that contains one or more logarithmic expressions with an unknown variable. For example, log_2(x) = 5 or log(x) + log(x - 3) = 1. Solving involves finding the value of the variable that makes the equation true.

How do you solve a basic logarithmic equation?

To solve a basic logarithmic equation like log_b(x) = c, rewrite it in exponential form: x = b^c. For example, log_2(x) = 5 becomes x = 2^5 = 32. Always check that the solution makes the argument of every logarithm positive.

What are extraneous solutions in logarithmic equations?

Extraneous solutions are values that emerge from the algebraic solving process but do not satisfy the original equation. They typically occur when a solution makes the argument of a logarithm zero or negative. Since log of a non-positive number is undefined, such solutions must be rejected.

Why must the argument of a logarithm be positive?

The argument must be positive because logarithms are defined only for positive real numbers. No real power of a positive base can produce zero or a negative number. For example, there is no real value of x such that 10^x = -5 or 10^x = 0.

When do you use logarithm properties to solve equations?

Use log properties when an equation has multiple log terms that need to be combined. The product rule combines log(M) + log(N) into log(MN). The quotient rule combines log(M) - log(N) into log(M/N). The power rule converts p*log(M) into log(M^p). After combining, rewrite in exponential form to solve.

How are logarithmic equations used in real life?

Logarithmic equations are used to calculate pH in chemistry, earthquake magnitudes on the Richter scale, sound levels in decibels, time for investments to grow to a target value, radioactive decay half-lives, and algorithm complexity in computer science.

Logarithmic Equations - Complete Study Guide

Last updated: March 2026 | AP - IB - GCSE - IGCSE - SAT

1. Introduction to Logarithmic Equations

What Are Logarithmic Equations?

A logarithmic equation is any equation that contains one or more logarithmic expressions in which the variable you need to find appears inside a logarithm, as the base of a logarithm, or alongside logarithmic terms. Examples include equations like log₂(x) = 5, or ln(x) + ln(x - 3) = 2, or log(x²) = 4.

In simpler terms, if you see "log" or "ln" in an equation with an unknown variable, you are looking at a logarithmic equation. Your task is to find the value (or values) of the variable that make the equation true.

Why Are Logarithmic Equations Important?

Logarithmic equations sit at the crossroads of three fundamental areas of mathematics: algebra, exponents, and logarithms. They combine the equation-solving skills you developed in algebra with your understanding of exponential and logarithmic relationships. Being able to solve them is essential because:

They appear on every major exam: AP Calculus, IB Mathematics, GCSE/IGCSE, SAT, and A-Level exams all include logarithmic equation questions.
They solve real-world problems: Finding how long an investment takes to reach a target value, calculating the time for radioactive material to decay to a safe level, or determining the pH of a chemical solution -- all require solving logarithmic equations.
They are essential for calculus: Many calculus problems require simplifying or solving logarithmic equations as an intermediate step.
They develop mathematical reasoning: Logarithmic equations require you to apply multiple skills together -- rewriting forms, applying rules, checking domain restrictions -- which builds deep problem-solving ability.

Why Students Find Logarithmic Equations Challenging

Students often struggle with logarithmic equations for three main reasons:

Multiple skills needed: You must be comfortable with logarithm definitions, exponent rules, logarithm properties, equation solving, and domain restrictions -- all in one problem.
Extraneous solutions: Unlike most algebra equations, logarithmic equations can produce "fake" answers that look correct but are actually invalid. You must check every answer, which is an unusual requirement for many students.
Unfamiliar notation: The log and ln notation still feels abstract to many students, making it harder to see what steps to take.

The good news: once you learn the systematic approach taught in this guide, logarithmic equations become very manageable. There are only a few basic types, and each follows a clear strategy.

Teacher Tip: The number one strategy for logarithmic equations is: get a single logarithm on one side, then convert to exponential form. If you remember nothing else, remember this.

2. Review of Logarithm Basics

Before solving logarithmic equations, let us make sure we have a solid foundation. If you are already confident with logs, skim this section as a quick refresher.

The Logarithm Definition

log_b(a) = c means b^c = a

A logarithm answers the question: "b raised to what power gives a?" The answer is c.

b = the base (must be positive and not equal to 1)
a = the argument (must be positive)
c = the exponent (the result -- can be any real number)

Quick Review Examples

Example 1 - log₂(8) = ?

Question: 2 to what power gives 8?

2¹ = 2, 2² = 4, 2³ = 8. Answer: 3

Example 2 - log₁₀(100) = ?

Question: 10 to what power gives 100?

10² = 100. Answer: 2

Example 3 - ln(e⁴) = ?

Question: ln and e are inverses, so they cancel: ln(e⁴) = 4

Common vs Natural Logarithms

Common logarithm: log(x) = log₁₀(x). Base 10. The "log" button on your calculator.
Natural logarithm: ln(x) = log_e(x). Base e (approximately 2.71828). The "ln" button on your calculator.

3. Connection Between Exponential Form and Logarithmic Form

The single most important technique for solving logarithmic equations is converting between logarithmic form and exponential form. These two forms say exactly the same thing in different notation.

Logarithmic Form: log_b(a) = c

Exponential Form: b^c = a

When you have a logarithmic equation in the form "log equals a number," convert to exponential form and the equation becomes much simpler to solve.

Conversion Example 1

Logarithmic: log₂(x) = 5

Exponential: 2⁵ = x, so x = 32

Conversion Example 2

Logarithmic: log₃(x + 1) = 2

Exponential: 3² = x + 1, so 9 = x + 1, so x = 8

Conversion Example 3

Logarithmic: ln(x) = 3

Exponential: e³ = x, so x = e³ (approximately 20.086)

Conversion Example 4

Logarithmic: log(x) = -2

Exponential: 10^-2 = x, so x = 1/100 = 0.01

4. Solving Basic Logarithmic Equations

A "basic" logarithmic equation has the form: one logarithm = a number. The strategy is always the same: convert to exponential form, then solve.

Example 1 - Solve log₂(x) = 4

Step 1 - Convert to exponential form: 2⁴ = x

Step 2 - Evaluate: x = 16

Step 3 - Check: log₂(16) = 4? Yes, because 2⁴ = 16. Valid.

Answer: x = 16

Example 2 - Solve log₁₀(x) = 3

Step 1 - Convert: 10³ = x

Step 2 - Evaluate: x = 1000

Step 3 - Check: log(1000) = 3? Yes. Valid.

Answer: x = 1000

Example 3 - Solve ln(x) = 2

Step 1 - Convert: e² = x

Step 2 - Evaluate: x = e² (approximately 7.389)

Step 3 - Check: ln(e²) = 2? Yes. Valid.

Answer: x = e² (approximately 7.389)

Example 4 - Solve log₅(x - 1) = 2

Step 1 - Convert: 5² = x - 1

Step 2 - Solve: 25 = x - 1, so x = 26

Step 3 - Check domain: x - 1 = 25 > 0. Valid.

Answer: x = 26

Example 5 - Solve log₄(2x + 3) = 3

Step 1 - Convert: 4³ = 2x + 3

Step 2 - Solve: 64 = 2x + 3 --> 2x = 61 --> x = 30.5

Step 3 - Check domain: 2(30.5) + 3 = 64 > 0. Valid.

Answer: x = 30.5

Teacher Tip: For basic logarithmic equations, the process is always: (1) isolate the log, (2) convert to exponential form, (3) solve the resulting equation, (4) check the domain. Master this four-step process and you will handle most exam questions.

5. Solving Logarithmic Equations Using Logarithm Properties

When a logarithmic equation has multiple log terms, you need to use logarithm properties to combine them into a single log before converting to exponential form. Here are the three essential rules.

The Three Key Rules

Product Rule: log_b(M * N) = log_b(M) + log_b(N)

Quotient Rule: log_b(M / N) = log_b(M) - log_b(N)

Power Rule: log_b(M^p) = p * log_b(M)

Strategy: Use these rules to combine multiple logs into one. Then rewrite the single-log equation in exponential form and solve.

Example 1 - Product Rule Equation

Solve: log(x) + log(x - 3) = 1

Step 1 - Identify the rule: Two logs being added with the same base (10). Use the Product Rule to combine.

Step 2 - Combine: log(x(x - 3)) = 1, which gives log(x² - 3x) = 1

Step 3 - Convert to exponential form: 10¹ = x² - 3x, so x² - 3x = 10

Step 4 - Solve the quadratic: x² - 3x - 10 = 0. Factor: (x - 5)(x + 2) = 0. So x = 5 or x = -2.

Step 5 - Check domain restrictions:

x = 5: log(5) is defined (5 > 0) and log(5 - 3) = log(2) is defined (2 > 0). VALID.
x = -2: log(-2) is undefined (-2 is not positive). REJECTED.

Answer: x = 5

Example 2 - Quotient Rule Equation

Solve: log₂(x) - log₂(x - 2) = 3

Step 1 - Identify the rule: Two logs being subtracted with the same base (2). Use the Quotient Rule.

Step 2 - Combine: log₂(x / (x - 2)) = 3

Step 3 - Convert to exponential form: 2³ = x / (x - 2), so 8 = x / (x - 2)

Step 4 - Solve: 8(x - 2) = x --> 8x - 16 = x --> 7x = 16 --> x = 16/7 (approximately 2.286)

Step 5 - Check domain:

x = 16/7: log₂(16/7) is defined (16/7 > 0). Check x - 2 = 16/7 - 14/7 = 2/7 > 0. VALID.

Answer: x = 16/7 (approximately 2.286)

Example 3 - Power Rule Equation

Solve: 2 log₃(x) = 4

Step 1 - Apply Power Rule in reverse: 2 log₃(x) = log₃(x²), so log₃(x²) = 4

Step 2 - Convert to exponential form: 3⁴ = x², so x² = 81

Step 3 - Solve: x = 9 or x = -9

Step 4 - Check domain:

x = 9: log₃(9) is defined. VALID.
x = -9: log₃(-9) is undefined. REJECTED.

Answer: x = 9

Example 4 - Combining Product and Quotient Rules

Solve: log(x) + log(x + 1) - log(2) = 1

Step 1 - Combine: log(x(x + 1) / 2) = 1

Step 2 - Convert: 10¹ = x(x + 1) / 2, so x(x + 1) = 20

Step 3 - Solve quadratic: x² + x - 20 = 0. Factor: (x + 5)(x - 4) = 0. So x = -5 or x = 4.

Step 4 - Check domain:

x = 4: log(4) defined, log(5) defined. VALID.
x = -5: log(-5) undefined. REJECTED.

Answer: x = 4

6. Domain Restrictions and Extraneous Solutions

This section addresses what is arguably the most important concept specific to logarithmic equations -- the fact that solving logarithmic equations can produce answers that are mathematically invalid.

Why the Argument Must Be Positive

Every logarithmic expression log_b(a) requires a > 0. There is no real exponent that makes a positive base produce zero or a negative number. Therefore, any solution to a logarithmic equation that makes any log argument zero or negative must be rejected.

What Are Extraneous Solutions?

An extraneous solution is a value that emerges from an algebraically correct solving process but does not satisfy the original equation. In logarithmic equations, extraneous solutions arise because:

When we combine logs using product or quotient rules, we sometimes change the domain of the equation.
When we square or factor, we introduce possibilities that did not exist in the original problem.

CRITICAL: Always Check Every Solution

After solving any logarithmic equation, substitute each answer back into the original equation and verify that every log argument is positive. If any argument is zero or negative, reject that solution.

Detailed Example with an Extraneous Solution

Solve: log₂(x) + log₂(x - 6) = 4

Step 1 - Combine (Product Rule): log₂(x(x - 6)) = 4

Step 2 - Convert: 2⁴ = x(x - 6), so x² - 6x = 16

Step 3 - Solve quadratic: x² - 6x - 16 = 0. Factor: (x - 8)(x + 2) = 0. So x = 8 or x = -2.

Step 4 - Check solutions in the ORIGINAL equation:

x = 8: log₂(8) + log₂(8 - 6) = log₂(8) + log₂(2) = 3 + 1 = 4. VALID.
x = -2: log₂(-2) is undefined. REJECTED (extraneous).

Answer: x = 8 only. The value x = -2 is extraneous.

Another Example - Both Solutions Could Be Extraneous

Solve: log₃(x - 4) + log₃(x + 4) = 2

Step 1: log₃((x - 4)(x + 4)) = 2, so log₃(x² - 16) = 2

Step 2: 3² = x² - 16, so 9 = x² - 16, so x² = 25

Step 3: x = 5 or x = -5

Step 4 - Check:

x = 5: (5 - 4) = 1 > 0, (5 + 4) = 9 > 0. VALID.
x = -5: (-5 - 4) = -9 < 0. REJECTED.

Answer: x = 5

Pro Tip: Before solving, write down the domain restrictions. For each log expression, set its argument > 0. This gives you a quick way to reject invalid solutions without fully substituting. For example, if you have log(x - 3), then x must be greater than 3. Any solution with x less than or equal to 3 is immediately invalid.

7. Solving More Advanced Logarithmic Equations

7.1 Equations with Logarithms on Both Sides

When both sides of the equation contain a single logarithm with the same base, use the one-to-one property:

If log_b(M) = log_b(N), then M = N

Example - Solve: log₅(2x + 1) = log₅(x + 4)

Step 1 - Apply one-to-one property: 2x + 1 = x + 4

Step 2 - Solve: x = 3

Step 3 - Check: 2(3) + 1 = 7 > 0, 3 + 4 = 7 > 0. VALID.

Answer: x = 3

7.2 Equations with Natural Logarithms

Solve: ln(x) + ln(x + 2) = ln(3)

Step 1 - Combine left side: ln(x(x + 2)) = ln(3)

Step 2 - One-to-one: x(x + 2) = 3, so x² + 2x - 3 = 0

Step 3 - Factor: (x + 3)(x - 1) = 0, so x = -3 or x = 1

Step 4 - Check:

x = 1: ln(1) = 0, ln(3) defined. 0 + ln(3) = ln(3). VALID.
x = -3: ln(-3) undefined. REJECTED.

Answer: x = 1

7.3 Equations Producing Quadratics

Solve: log₂(x² - 4x) = 3

Step 1 - Convert: 2³ = x² - 4x, so x² - 4x = 8

Step 2 - Solve: x² - 4x - 8 = 0. Quadratic formula: x = (4 +/- sqrt(16 + 32)) / 2 = (4 +/- sqrt(48)) / 2 = (4 +/- 4sqrt(3)) / 2 = 2 +/- 2sqrt(3)

Step 3 - Evaluate: x = 2 + 2sqrt(3) (approximately 5.46) or x = 2 - 2sqrt(3) (approximately -1.46)

Step 4 - Check domain: The argument is x² - 4x.

x = 2 + 2sqrt(3): argument = 8 > 0. VALID.
x = 2 - 2sqrt(3): argument = 8 > 0 (same value!). VALID.

Answer: x = 2 + 2sqrt(3) and x = 2 - 2sqrt(3) (both valid!)

7.4 Equations Requiring Substitution

Solve: (log(x))² - 3 log(x) + 2 = 0

Step 1 - Let u = log(x): u² - 3u + 2 = 0

Step 2 - Factor: (u - 1)(u - 2) = 0, so u = 1 or u = 2

Step 3 - Back-substitute:

u = 1: log(x) = 1, so x = 10¹ = 10
u = 2: log(x) = 2, so x = 10² = 100

Step 4 - Check: Both x = 10 and x = 100 are positive. VALID.

Answer: x = 10 or x = 100

7.5 Equations with Different Bases

Solve: log₂(x) = log₄(3x + 4)

Step 1 - Convert to same base: Since 4 = 2², use change of base: log₄(3x + 4) = log₂(3x + 4) / log₂(4) = log₂(3x + 4) / 2

Step 2 - Substitution: Let y = log₂(x). Then log₂(3x + 4) / 2 = y, so log₂(3x + 4) = 2y

Step 3 - From log₂(x) = y: x = 2^y. From log₂(3x + 4) = 2y: 3x + 4 = 2^2y = (2^y)² = x²

Step 4 - Solve: x² = 3x + 4 --> x² - 3x - 4 = 0 --> (x - 4)(x + 1) = 0 --> x = 4 or x = -1

Step 5 - Check: x = 4 is valid (positive). x = -1 is rejected (log₂(-1) undefined).

Answer: x = 4

8. Logarithmic Equations and Exponential Equations

Logarithmic equations and exponential equations are two sides of the same coin. Understanding this relationship is essential because many problems require converting between the two types.

The Core Relationship

Exponential equation: b^x = a

Logarithmic equivalent: x = log_b(a)

When you cannot solve an exponential equation by matching bases, you take a logarithm of both sides. When you cannot solve a logarithmic equation directly, you convert to exponential form. They are solving strategies that complement each other.

Side-by-Side Comparison

Exponential Equation	Solving Method	Logarithmic Equivalent
2^x = 16	Match bases: 2^x = 2⁴	x = log₂(16) = 4
5^x = 200	Take log: x = log(200)/log(5)	x = log₅(200) = 3.292
e^x = 7	Take ln: x = ln(7)	x = ln(7) = 1.946
3^2x+1 = 81	Match: 3^2x+1 = 3⁴	2x + 1 = 4, x = 1.5

Example - Converting Between Types

Exponential problem: How long does it take $1000 to grow to $2000 at 6% annual interest?

Equation: 1000 * (1.06)^t = 2000, so (1.06)^t = 2

Take log of both sides: t * log(1.06) = log(2)

Solve: t = log(2) / log(1.06) = 0.3010 / 0.0253 = 11.90 years

This shows how a logarithmic equation naturally arises from solving an exponential problem.

9. Graphical Understanding of Logarithmic Equations

Understanding logarithmic equations graphically gives you powerful visual intuition about solutions.

Solving Graphically

The equation log_b(x) = c can be understood as: "Where does the graph y = log_b(x) intersect the horizontal line y = c?"

Key observations about the graph of y = log_b(x) (when b > 1):

Domain: x > 0 only. The graph exists only to the right of the y-axis, which is why log arguments must be positive.
Always passes through (1, 0): Because log_b(1) = 0 for any base.
Always passes through (b, 1): Because log_b(b) = 1.
Strictly increasing: The graph always rises (for b > 1), meaning every horizontal line y = c intersects it in exactly one point. This is why basic logarithmic equations have exactly one solution.
Vertical asymptote at x = 0: The graph approaches but never touches the y-axis.

Why Some Equations Have More Solutions

When you have equations like log(x²) = 2, the squared argument creates a quadratic that can produce two solutions. Graphically, you are solving log(u) = 2 where u = x², and the parabola u = x² gives two x-values for each positive u.

When There Is No Solution

If a logarithmic equation demands a negative argument (for example, after simplification you get log₂(x) = 3 with an additional constraint forcing x < 0), the equation has no real solution. Graphically, the curve y = log_b(x) never reaches the region x < 0.

10. Real-Life Applications of Logarithmic Equations

Logarithmic equations are not just abstract exam problems. They arise naturally whenever you need to "undo" an exponential relationship to find time, rate, or quantity.

10.1 pH Calculations (Chemistry)

The pH of a solution is defined as pH = -log[H+], where [H+] is the hydrogen ion concentration. To find the concentration from a given pH, you solve the logarithmic equation: -log[H+] = pH, so [H+] = 10^-pH. For example, a solution with pH = 4.5 has [H+] = 10^-4.5 = 3.16 * 10^-5 mol/L.

10.2 Earthquake Magnitude (Richter Scale)

The Richter scale magnitude is M = log(I/I₀), where I is the earthquake intensity and I₀ is a reference intensity. To find the intensity of a magnitude 7 earthquake: log(I/I₀) = 7, so I = I₀ * 10⁷. To compare a magnitude 7 and magnitude 5 earthquake: their intensity ratio is 10⁷ / 10⁵ = 10² = 100 times stronger.

10.3 Sound Intensity (Decibels)

Sound level in decibels is L = 10 log(I/I₀). If a speaker produces 90 dB, what is the intensity? Solve: 10 log(I/I₀) = 90 --> log(I/I₀) = 9 --> I/I₀ = 10⁹ --> I = 10⁹ * I₀. This is a direct logarithmic equation.

10.4 Finance - Time to Reach a Target Value

With compound interest A = P(1 + r)^t, finding when an investment reaches a target requires solving a logarithmic equation. For example: when does $5000 at 8% reach $10,000?

10000 = 5000(1.08)^t --> 2 = (1.08)^t --> t = ln(2)/ln(1.08) = 0.6931/0.0770 = 9.0 years

10.5 Radioactive Decay (Science)

Radioactive decay follows N = N₀ * e^-kt. To find the half-life, set N = N₀/2: 1/2 = e^-kt --> ln(1/2) = -kt --> t = ln(2)/k. Every half-life calculation involves solving a logarithmic equation.

10.6 Computer Science - Algorithm Analysis

Binary search runs in O(log₂(n)) steps. If you want binary search to find an item in at most 20 steps, solve: log₂(n) = 20 --> n = 2²⁰ = 1,048,576. So you can search through over a million sorted items in just 20 steps.

Practical Example - Population Growth

Problem: A city's population grows according to P = 50000 * e^0.03t. When will the population reach 100,000?

Set up: 100000 = 50000 * e^0.03t

Step 1: Divide: 2 = e^0.03t

Step 2: Take ln: ln(2) = 0.03t

Step 3: Solve: t = ln(2) / 0.03 = 0.6931 / 0.03 = 23.1 years

11. Common Student Mistakes

These are the most frequent errors seen in twenty years of teaching logarithmic equations. Study them carefully.

Mistake 1: Forgetting to Check Domain Restrictions

Wrong: Solving log(x) + log(x - 3) = 1 and accepting both x = 5 and x = -2.

Correct: x = -2 makes log(-2) undefined. It must be rejected. Only x = 5 is valid.

Always check that every log argument is positive with your final answer.

Mistake 2: Thinking log(a + b) = log(a) + log(b)

Wrong: log(x + 5) = log(x) + log(5)

Correct: The product rule states log(a * b) = log(a) + log(b). Addition inside the log does NOT distribute. There is no rule for log(a + b).

Mistake 3: Applying Log Rules to the Wrong Bases

Wrong: Combining log₂(x) + log₃(x) into a single log.

Correct: The product, quotient, and power rules only work when logs have the same base. Different bases require the change of base formula first.

Mistake 4: Not Fully Solving After Converting to Exponential Form

Wrong: Solving log₂(x - 3) = 4 and writing x - 3 = 16 as the final answer.

Correct: You must finish solving: x - 3 = 16, so x = 19. Then check: 19 - 3 = 16 > 0. Valid.

Mistake 5: Confusing log(x)² with log(x²)

log(x)² means [log(x)]² -- you are squaring the result of the log.

log(x²) means log of x-squared -- you can apply the power rule: 2 log(x).

These are different expressions! Be careful with parentheses and notation.

Mistake 6: Algebra Errors After Conversion

Issue: Students correctly convert to exponential form but then make sign errors, distribution errors, or factoring mistakes in the resulting algebraic equation.

Fix: Slow down when solving the algebra. Show every step. Double-check by substituting your final answer back into the original logarithmic equation.

12. Exam Tips and Problem-Solving Strategies

Strategy 1: Identify the Equation Type First

Before doing anything, classify the equation:

Single log = number --> Convert to exponential form directly.
Multiple logs on one side --> Use product/quotient/power rules to combine, then convert.
Log on both sides --> Use one-to-one property (if same base) or convert to same base.
Quadratic in log --> Substitute u = log(x), solve the quadratic, then back-substitute.

Strategy 2: Write Domain Restrictions Before Solving

Before you begin solving, look at every log expression in the equation and write down its domain requirement. For example, if you have log(x - 3) + log(x + 1) = 2, immediately write: x - 3 > 0 (x > 3) AND x + 1 > 0 (x > -1). The stricter condition wins: x > 3. Any answer less than or equal to 3 is automatically rejected.

Strategy 3: When in Doubt, Convert to Exponential Form

The most reliable technique: get a single log on one side, convert to exponential form, and solve the resulting equation. This works for the vast majority of logarithmic equations on exams.

Strategy 4: Always Verify by Substitution

After finding solutions, substitute each one back into the original equation (not your simplified version). Verify that: (a) all log arguments are positive, and (b) the equation is actually satisfied. This takes 30 seconds and can save you from losing marks for extraneous solutions.

Strategy 5: Use Change of Base When Bases Differ

If the equation has logarithms with different bases, use the change of base formula to convert everything to one base (usually 10 or e). Then combine and solve normally.

Strategy 6: Show All Working for Full Marks

Write each step clearly: state which rule you are using, show the exponential conversion, solve the algebra, and explicitly state which solutions are valid and which are rejected. Examiners award method marks generously for logarithmic equations.

13. Practice Problems

Test yourself with these 15 problems. Work each one on paper first, then click "Show Solution" to check your answer.

Question 1

Solve: log₃(x) = 4

Convert: 3⁴ = x, so x = 81

Check: log₃(81) = 4. Valid.

Question 2

Solve: log(x) = -1

Convert: 10^-1 = x, so x = 0.1

Check: log(0.1) = -1. Valid.

Question 3

Solve: ln(x) = 4

Convert: e⁴ = x, so x = e⁴ (approximately 54.598)

Question 4

Solve: log₂(3x - 1) = 5

Convert: 2⁵ = 3x - 1 --> 32 = 3x - 1 --> 3x = 33 --> x = 11

Check: 3(11) - 1 = 32 > 0. Valid.

Question 5

Solve: log(x) + log(x + 3) = 1

Product Rule: log(x(x + 3)) = 1 --> 10¹ = x² + 3x --> x² + 3x - 10 = 0

Factor: (x + 5)(x - 2) = 0 --> x = -5 or x = 2

Check: x = -5 gives log(-5), undefined. REJECTED. x = 2: log(2) + log(5) = log(10) = 1. VALID.

Answer: x = 2

Question 6

Solve: log₃(x + 2) - log₃(x) = 1

Quotient Rule: log₃((x + 2)/x) = 1 --> 3¹ = (x + 2)/x --> 3x = x + 2 --> 2x = 2 --> x = 1

Check: log₃(3) - log₃(1) = 1 - 0 = 1. Valid.

Question 7

Solve: 2 ln(x) = 6

Divide by 2: ln(x) = 3 --> x = e³ = approximately 20.086

Check: 2 ln(e³) = 2(3) = 6. Valid.

Question 8

Solve: log₄(x) + log₄(x - 6) = 2

Product Rule: log₄(x(x - 6)) = 2 --> 4² = x² - 6x --> x² - 6x - 16 = 0

Factor: (x - 8)(x + 2) = 0 --> x = 8 or x = -2

Check: x = -2 gives log₄(-2), undefined. REJECTED. x = 8: both arguments positive. VALID.

Answer: x = 8

Question 9

Solve: log₅(2x + 3) = log₅(x + 7)

One-to-one: 2x + 3 = x + 7 --> x = 4

Check: 2(4) + 3 = 11 > 0, 4 + 7 = 11 > 0. Valid.

Question 10

Solve: log(x²) = 4

Convert: 10⁴ = x² --> x² = 10000 --> x = 100 or x = -100

Check: log((-100)²) = log(10000) = 4. Both VALID (the argument x² is positive for both).

Answer: x = 100 or x = -100

Question 11

Solve: log₂(x - 1) + log₂(x + 1) = 3

Product: log₂((x-1)(x+1)) = 3 --> log₂(x² - 1) = 3 --> x² - 1 = 8 --> x² = 9 --> x = 3 or x = -3

Check: x = -3 gives log₂(-4), undefined. REJECTED. x = 3: both positive. VALID.

Answer: x = 3

Question 12

Solve: ln(x + 2) = ln(5)

One-to-one: x + 2 = 5 --> x = 3

Check: ln(5) = ln(5). Valid.

Question 13

Solve: (log(x))² - log(x) - 2 = 0

Let u = log(x): u² - u - 2 = 0 --> (u - 2)(u + 1) = 0 --> u = 2 or u = -1

Back-substitute: log(x) = 2 --> x = 100. log(x) = -1 --> x = 0.1.

Check: Both positive. VALID.

Answer: x = 100 or x = 0.1

Question 14

Solve: log₂(x) = 1 + log₂(x - 2)

Rearrange: log₂(x) - log₂(x - 2) = 1

Quotient: log₂(x/(x - 2)) = 1 --> 2¹ = x/(x - 2) --> 2(x - 2) = x --> 2x - 4 = x --> x = 4

Check: log₂(4) = 2, log₂(2) = 1. 2 = 1 + 1. Valid.

Question 15

Solve: log₃(x² + 3x) = log₃(10)

One-to-one: x² + 3x = 10 --> x² + 3x - 10 = 0 --> (x + 5)(x - 2) = 0

x = -5 or x = 2

Check: x = -5: (-5)² + 3(-5) = 25 - 15 = 10 > 0. VALID.

x = 2: (2)² + 3(2) = 4 + 6 = 10 > 0. VALID.

Answer: x = -5 or x = 2

Interactive Logarithm Helper

Logarithm Equation Helper

Enter a base and argument to evaluate the logarithm and see the equivalent exponential form.

Base (b):

Argument (a):

14. Summary of Key Points

Concept / Rule	Formula	Key Idea
Definition	log_b(a) = c means b^c = a	"b to what power gives a?"
Solving Strategy	Single log = number --> exponential form	Most reliable technique
Product Rule	log(M) + log(N) = log(M*N)	Addition of logs = log of product
Quotient Rule	log(M) - log(N) = log(M/N)	Subtraction of logs = log of quotient
Power Rule	p * log(M) = log(M^p)	Coefficient becomes exponent inside log
One-to-One Property	log_b(M) = log_b(N) --> M = N	Same log, same base --> arguments equal
Change of Base	log_b(a) = log(a) / log(b)	Convert to base 10 or e for calculator
Domain Restriction	Argument must be > 0	Always check every log argument
Extraneous Solutions	Solutions making argument non-positive	Must be rejected after checking
Verification	Substitute back into original equation	Essential for every logarithmic equation

Final Reminders

Get a single log on one side, then convert to exponential form.
Use product, quotient, and power rules to combine multiple logs first.
If logs are on both sides with the same base, use the one-to-one property.
Always check domain -- reject solutions where any log argument is not positive.
log(a + b) is NOT equal to log(a) + log(b). There is no sum rule!
Show all working for maximum marks on exams.
When the equation produces a quadratic, expect to check both roots -- one may be extraneous.

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