Logarithmic Properties - Complete Study Guide

Last updated: March 2026 | Algebra 2, Pre-Calculus, AP, IB, GCSE

1. Introduction to Logarithmic Properties

Welcome to the core engine of logarithmic mathematics. If logarithms are the tools we use to solve exponential problems, then logarithmic properties are the instruction manual for how those tools operate.

What Are Logarithmic Properties?

Logarithmic properties are a set of mathematical rules that govern how logarithms interact with arithmetic operations like multiplication, division, and exponentiation. The three most famous rules -- the Product, Quotient, and Power properties -- dictate how we can tear complex logarithms apart (expanding) or squeeze multiple logarithms together (condensing).

Why Are They Important in Algebra?

In algebra, we constantly need to manipulate equations to isolate a variable. Without logarithmic properties, we would hit a brick wall anytime a variable was buried inside a complex logarithm, or stuck across multiple logarithmic terms. These properties give us the power to:

  1. Simplify monstrous expressions: Turning a frightening expression into manageable, bite-sized pieces.
  2. Solve equations: If an equation has four different logarithms in it, you cannot solve it directly. You must use the properties to condense them into a single logarithm first.
  3. Calculate without calculators: Before modern digital calculators, mathematicians used tables of logarithm properties to turn tedious multiplication of massive numbers into simple addition.

Why Do Students Find Them Confusing?

If you feel a bit dizzy looking at log properties for the first time, don't worry. Students commonly struggle because these rules act like "funhouse mirrors" of normal arithmetic:

  • Normal arithmetic: Multiplication is multiplication.
  • Logarithm arithmetic: Multiplication inside the log becomes addition outside the log.
  • Normal arithmetic: Division is division.
  • Logarithm arithmetic: Division inside the log becomes subtraction outside.

It takes practice to rewire your brain to accept that logs translate higher-level operations (multiplication/division) down to lower-level operations (addition/subtraction). But once it clicks, it is remarkably consistent.

Real-Life Uses

Whether you are calculating the pH of a new chemical compound, adjusting the decibel level on a soundboard for a concert, analyzing Richter scale data for tectonic activity, or predicting compound interest on a mortgage, logarithmic properties are happening under the mathematical hood. They are exactly what allows us to model these massive-scale phenomena gracefully.

2. Review of Logarithm Basics

Before jumping into the properties, we must speak the language of logarithms fluently. A logarithm is fundamentally asking a specific question.

logb(a) = c   means exactly   bc = a

"Base 'b' raised to what power 'c' gives me the argument 'a'?"

  • Base (b): The number at the bottom. It must be positive and not equal to 1.
  • Argument (a): The number inside the parentheses. It must be strictly positive.
  • Result/Exponent (c): The answer to the log. This is the exponent you needed to find.

Common and Natural Logs

  • Common Logarithm: When you see log(x) with no base written, it implies base 10. (log10(x)).
  • Natural Logarithm: When you see ln(x), it implies base e, where e is Euler's number (approx 2.718). (loge(x)).

Basic Example 1: log2(8)

Question: 2 to what power gives 8?

Thinking: 21=2, 22=4, 23=8.

Answer: log2(8) = 3

Basic Example 2: log(100)

Question: (Base 10 hidden) 10 to what power gives 100?

Thinking: 102 = 100.

Answer: log(100) = 2

Basic Example 3: ln(e4)

Question: (Base e hidden) e to what power gives e4?

Thinking: To get e4, I need an exponent of 4.

Answer: ln(e4) = 4

3. Connection Between Exponents and Logarithms

Logarithms and exponents are inverse functions. In plain English, they undo each other. The ability to smoothly translate a statement from exponential form into logarithmic form (and vice versa) is the most critical survival skill in this topic.

Exponential Form: 52 = 25
Logarithmic Form: log5(25) = 2

Notice the pattern: The base of the exponent (5) becomes the base of the log. The result (25) becomes the argument. The exponent (2) moves to the other side of the equals sign to become the answer.

Converting from Exponential to Logarithmic

Example A: 23 = 8

The base is 2. The exponent is 3. The result is 8.

Translation: log2(8) = 3

Example B: 10-2 = 0.01

The base is 10. The exponent is -2. The result is 0.01.

Translation: log10(0.01) = -2, or simply log(0.01) = -2

Converting from Logarithmic to Exponential

Example C: log5(1) = 0

The base is 5. The result/exponent is 0. The argument is 1.

Translation: 50 = 1

Example D: ln(x) = 5

The base is e (hidden implicitly). The exponent is 5. The argument is x.

Translation: e5 = x

💡 The "Circle" Trick: To convert logb(a) = c, start at the base b, draw a circular arrow over the equals sign to the c, and then circle back to the a. "b raised to c equals a".

4. Main Logarithmic Properties

Now that you know the language, let's learn the rules. These six properties are the foundation of all logarithmic algebra. Memorize them, understand them, and practice them.

4.1 Product Property

logb(M * N) = logb(M) + logb(N)

The logarithm of a product is the sum of the logarithms.

Explanation: When you have multiplication inside the argument of a logarithm, you can split it into two separate logarithms joined by addition. The base remains the same.

Example:

log2(8 * 4) = log2(8) + log2(4)

Let's check if this is true: log2(32) = 5, and 3 + 2 = 5. It works perfectly!

4.2 Quotient Property

logb(M / N) = logb(M) - logb(N)

The logarithm of a quotient is the difference of the logarithms.

Explanation: Division inside the logarithm splits into two separate logarithms joined by subtraction. The numerator goes first (positive), and the denominator goes second (negative).

Example:

log3(27 / 9) = log3(27) - log3(9)

Check: log3(3) = 1, and 3 - 2 = 1. Correct.

4.3 Power Property

logb(Mp) = p * logb(M)

The exponent of an argument can be moved to the front as a multiplier.

Explanation: If the argument has an exponent, you can "drop it down" in front of the logarithm to multiply the whole expression. This is perhaps the most powerful rule for solving equations.

Example:

log10(1003) = 3 * log10(100)

Check: log(1,000,000) = 6, and 3 * 2 = 6. Correct.

4.4 Log of 1 Property

logb(1) = 0

The logarithm of 1 to any valid base is always 0.

Explanation: Why? Because any non-zero number raised to the power of 0 equals 1. For example: 50 = 1, so log5(1) = 0.

4.5 Log of Base Property

logb(b) = 1

The logarithm of the base itself is always 1.

Explanation: Why? Because any number raised to the 1st power is itself. For example: 71 = 7, so log7(7) = 1.

4.6 Change of Base Formula

logb(a) = logc(a) / logc(b)

(Usually used as: logb(a) = log(a) / log(b)   or   ln(a) / ln(b))

Explanation: If your calculator only has log (base 10) and ln (base e) buttons, how do you calculate log2(7)? You use this formula. You choose a new base (like 10), take the log of the argument, and divide it by the log of the old base.

Example:

log2(7) = log(7) / log(2) ≈ 0.845 / 0.301 ≈ 2.807

5. Why These Properties Work

Do not just memorize these rules blindly! Logarithms are exponents. Therefore, logarithmic properties are just the laws of exponents in reverse.

Proof of the Product Property

Let's remember our exponent rule for multiplication: bx * by = bx+y. When you multiply numbers with the same base, you add the exponents.

Now, let's translate this to logarithms.

Let x = logb(M) and y = logb(N).

Converting to exponential form gives: bx = M and by = N.

What happens if we multiply M and N?
M * N = bx * by = bx+y

So, M * N = bx+y.

If we convert this back to logarithmic form, we get:
logb(M * N) = x + y

Substitute back what x and y were originally:
logb(M * N) = logb(M) + logb(N)

Conclusion: Multiplication inside the log becomes addition outside because multiplying exponential numbers means adding their exponents.

Proof of the Quotient Property

Exponent rule for division: bx / by = bx-y. When you divide numbers with the same base, you subtract the exponents.

Following the same logic as above:
M / N = bx / by = bx-y

Convert to log form: logb(M / N) = x - y

Substitute back: logb(M / N) = logb(M) - logb(N)

Proof of the Power Property

Exponent rule for powers: (bx)p = bx*p. When raising a power to a power, you multiply the exponents.

Let x = logb(M), so bx = M.

Now, raise both sides to the power of p:
(bx)p = Mp
bpx = Mp

Convert to log form: logb(Mp) = p * x

Substitute back: logb(Mp) = p * logb(M)

6. Simplifying Logarithmic Expressions

A common question is to take a single, complex logarithmic expression and simplify it by breaking it down. This relies heavily on the Product, Quotient, and Power properties in combination.

Example 1: Simplify log2(8x)

Identify property: We have multiplication (8 times x) inside the log. Use the Product Property.

log2(8 * x) = log2(8) + log2(x)

Simplify further: Can we evaluate log2(8)? Yes, 23 = 8.

= 3 + log2(x)

Restriction: x must be positive (x > 0).

Example 2: Simplify log3(27x2)

Identify property: We have multiplication (27 times x2) inside the log. Use the Product Property first.

log3(27 * x2) = log3(27) + log3(x2)

Next steps: Evaluate log3(27), which is 3. Then use the Power Property on the x2 term to bring the 2 to the front.

= 3 + 2 * log3(x)

Example 3: Simplify log(x / 5)

Identify property: Division inside the argument. Use the Quotient Property.

log(x / 5) = log(x) - log(5)

Example 4: Simplify ln(e7)

Identify property: Power Property.

ln(e7) = 7 * ln(e)

Simplify further: Recall the Log of Base property. The base of ln is e, so ln(e) = 1.

= 7 * 1 = 7

Alternative logic: Since ln and e are inverses, they cancel each other out directly, leaving just the exponent 7.

Example 5: Simplify log10(1000a3 / b)

Identify property: This has multiplication, division, and a power. Start with the "biggest" operation: division (Quotient Property).

= log(1000a3) - log(b)

Next step: Use Product Property on the first term.

= (log(1000) + log(a3)) - log(b)

Final step: Evaluate log(1000) which is 3, and use Power Property on a3.

= 3 + 3log(a) - log(b)

7. Expanding and Condensing Logarithmic Expressions

This is where your property knowledge gets put to the test. Expanding means tearing a single complex logarithm apart into many simple logarithms. Condensing means squishing multiple simple logarithms back together into a single, compact logarithm.

Expanding Logarithms

Goal: Turn one logarithm into a sum or difference of simpler logarithms, and bring all exponents down to the front.

Order of operations for expanding: (1) Handle multiplication and division first (Product and Quotient Rules). (2) Handle exponents last (Power Rule).

Expand: log2(4x3 / y)

Step 1 (Quotient Rule): Break the fraction apart.
= log2(4x3) - log2(y)

Step 2 (Product Rule): Break the multiplication (4 times x3) apart. Note: Keep the minus sign on the y term.
= log2(4) + log2(x3) - log2(y)

Step 3 (Power Rule): Bring the exponent 3 down to the front.
= log2(4) + 3 log2(x) - log2(y)

Step 4 (Simplify): Evaluate log2(4) = 2.
Final Answer: 2 + 3 log2(x) - log2(y)

Expand: ln(a2b)

Step 1 (Product Rule): Break the multiplication.
= ln(a2) + ln(b)

Step 2 (Power Rule): Bring the exponent down.
Final Answer: 2 ln(a) + ln(b)

Condensing Logarithms

Goal: Combine multiple logarithms into one single logarithm. (Requirement: The bases must be exactly the same!)

Order of operations for condensing: (1) Handle coefficients first (Power Rule in reverse). (2) Handle addition and subtraction left-to-right (Product and Quotient Rules).

Condense: log(x) + log(y)

Step 1 (Product Rule backwards): Addition outside the logs becomes multiplication inside a single log.
Final Answer: log(x * y) or just log(xy)

Condense: 2 log3(a) - log3(b)

Step 1 (Power Rule backwards): Move the coefficient 2 up to become an exponent.
= log3(a2) - log3(b)

Step 2 (Quotient Rule backwards): Subtraction outside becomes division inside.
Final Answer: log3(a2 / b)

Condense: 3 ln(x) + ln(y) - 2 ln(z)

Step 1 (Power Rule backwards): Move all coefficients up.
= ln(x3) + ln(y) - ln(z2)

Step 2 (Product and Quotient backwards): Addition goes in the numerator, subtraction goes in the denominator.
Final Answer: ln( (x3y) / z2 )

8. Domain Restrictions and Validity

There is a golden rule in logarithms that you violate at your own peril:

You can NEVER take the logarithm of zero or a negative number.
The argument must ALWAYS be strictly greater than zero. (Argument > 0)

Why Can't the Argument Be Negative or Zero?

Let's pretend we want to solve log2(-4) = x. If we convert this to exponential form, it becomes 2x = -4.

Is there any power you can raise positive 2 to that will result in negative 4? No. If x is positive, 2x is positive. If x is zero, 20 = 1. If x is negative, 2-x creates a fraction (like 2-2 = 1/4), but it is still positive.

Because there is no real number that satisfies this, the expression is undefined in the real number system. (It requires complex numbers, which are beyond the scope of high school algebra).

Identifying Restrictions

To find the domain (the valid x-values), set the argument strictly greater than zero and solve the inequality.

Find the domain of log(x - 2)

Set argument > 0: x - 2 > 0

Solve: x > 2

Answer: The domain is all real numbers greater than 2. Interval notation: (2, ∞)

Find the domain of log(3x + 1)

Set argument > 0: 3x + 1 > 0

Solve: 3x > -1x > -1/3

Answer: Domain is x > -1/3. Interval notation: (-1/3, ∞)

Find the domain of log( (x + 2) / (x - 5) )

This is trickier. For a fraction to be positive, both numerator and denominator must have the same sign (both positive OR both negative).

  • Case 1 (both positive): x + 2 > 0 (so x > -2) AND x - 5 > 0 (so x > 5). Overlap: x > 5.
  • Case 2 (both negative): x + 2 < 0 (so x < -2) AND x - 5 < 0 (so x < 5). Overlap: x < -2.

Answer: The domain is x < -2 or x > 5. Interval notation: (-∞, -2) ∪ (5, ∞)

9. Using Logarithmic Properties in Equations

Properties aren't just for simplifying; they are the key to unlocking solving logarithmic equations. The primary strategy makes use of condensing: if you have multiple logs on one side, condense them into a single log, then convert to exponential form to solve algebraically.

Step-by-Step Solving Examples

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

Step 1: Condense. Two logs added means multiply their arguments (Product Rule).

log(x * (x - 3)) = 1
log(x2 - 3x) = 1

Step 2: Convert to Exponential Form. The base is 10 (hidden).

101 = x2 - 3x
10 = x2 - 3x

Step 3: Solve Algebraic Equation. This is a quadratic. Move everything to one side.

x2 - 3x - 10 = 0
Factor: (x - 5)(x + 2) = 0
Possible solutions: x = 5 or x = -2

Step 4: Check Domain Restrictions. Substitute back into the original equation.

  • If x = 5: log(5) + log(2) = 1. Both arguments are positive. VALID.
  • If x = -2: log(-2) yields undefined. REJECTED.

Final Answer: x = 5

Solve: 2 log2(x) = 6

Step 1: Simplify/Condense. Use the Power Rule backwards to move the 2 inside. (Alternatively, just divide both sides by 2 first).

Power Rule: log2(x2) = 6

Step 2: Convert to Exponential Form.

26 = x2
64 = x2

Step 3: Solve Algebraic Equation. Square root both sides.

x = 8 or x = -8

Step 4: Check Domain Restrictions. Original argument must be positive.

  • If x = 8: log2(8) is valid. VALID.
  • If x = -8: log2(-8) is undefined. REJECTED.

Final Answer: x = 8

Solve: log3(x) - log3(x - 2) = 2

Step 1: Condense. Subtraction means divide arguments (Quotient Rule).

log3(x / (x - 2)) = 2

Step 2: Convert to Exponential Form.

32 = x / (x - 2)
9 = x / (x - 2)

Step 3: Solve Algebraic Equation. Multiply both sides by denominator.

9(x - 2) = x
9x - 18 = x
8x = 18
x = 18 / 8 = 9/4 = 2.25

Step 4: Check Domain Restrictions.

If x = 2.25: log3(2.25) is valid. log3(2.25 - 2) = log3(0.25) is valid. VALID.

Final Answer: x = 9/4 (or 2.25)

Solve: ln(x2) = 4

Step 1: Convert to Exponential Form. The base is e (hidden).

e4 = x2

Step 2: Solve Algebraic Equation. Square root both sides.

x = +√(e4) or x = -√(e4)
x = e2 or x = -e2

Step 3: Check Domain Restrictions. The original argument is x2.

  • If x = e2: (e2)2 = e4, which is positive. VALID.
  • If x = -e2: (-e2)2 = e4, which is also positive. VALID.

Final Answer: x = e2 and x = -e2 (Both are solutions!)

10. Common Logarithm vs Natural Logarithm

Do properties change if the base changes? NO. The properties of logarithms are universal across every positive real base (excluding 1). This specifically means that the rules for Common Logs (base 10) and Natural Logs (base e) look identical structurally.

  • Common Log (log): Product: log(MN) = log(M) + log(N). Quotient: log(M/N) = log(M) - log(N). Power: log(Mp) = p log(M).
  • Natural Log (ln): Product: ln(MN) = ln(M) + ln(N). Quotient: ln(M/N) = ln(M) - ln(N). Power: ln(Mp) = p ln(M).

You may use these symbols interchangeably in the property laws without modifying the logic whatsoever.

11. Real-Life Applications of Logarithmic Properties

Logarithmic properties are indispensable across diverse fields because they transform exponential growth (or decay) into linear, manageable datasets.

Chemistry and pH

The pH scale measures acidity: pH = -log[H+], where [H+] is hydrogen ion concentration. Because concentration spans many powers of 10, the logarithm condenses the massive range into small, readable numbers (like pH 4 to pH 8). If concentration increases by a factor of 10, the pH decreases by exactly 1 unit due to the power rule of logarithms.

Earthquake Magnitude (Richter Scale)

Magnitude M = log(I / I0). The quotient property separates the intensity I from the baseline I0. This is why a magnitude 7 earthquake isn't "one more" than a magnitude 6; it's exactly 10 times more intense!

Sound Intensity (Decibels)

Sound level L = 10 * log(I / I0). The human ear perceives sound logarithmically, not linearly. If an amplifier doubles its physical power output, the perceived volume does not double; it increases by about 3 decibels. We use log properties to design audio equipment matching human perception.

Finance and Exponential Growth

In the compound interest formula A = P(1 + r)t, if you want to find out how long (time t) it takes an investment to double, you must take the logarithm of both sides. log(2P/P) = log(1+r)t. The Power Property immediately pulls the unknown time variable t out of the exponent so it can be solved algebraically.

Computer Science Algorithms

Data sorting and searching algorithms (like Binary Search or Merge Sort) divide datasets in half repeatedly. Their efficiency is measured in "Big O" notation as O(n log n) or O(log n). Log properties help computer scientists calculate peak processing limits for billions of data points.

12. Common Student Mistakes

Logarithmic properties are strict. If you bend the rules even slightly, the math breaks. Here are the most frequent errors students make on tests.

Mistake 1: The "Fake" Sum Rule: log(a + b) = log(a) + log(b)

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

Correct: There is NO property that allows you to break apart addition inside a logarithm. log(x + 5) cannot be simplified further. The Product Property only applies to multiplication inside the log: log(x * 5) = log(x) + log(5).

Mistake 2: The "Fake" Difference Rule: log(a - b) = log(a) / log(b)

Wrong: log(10 - 2) = log(10) / log(2)

Correct: Subtraction inside the log cannot be broken apart. log(10 - 2) is simply log(8). Division inside the log becomes subtraction outside: log(10 / 2) = log(10) - log(2).

Mistake 3: Misapplying the Power Rule

Wrong: (log(x))3 = 3 log(x)

Correct: The Power Property only applies when the exponent is on the argument only. log(x3) = 3 log(x). If the entire logarithm is raised to a power, like (log(x))3, you cannot move the 3 to the front.

Mistake 4: Condensing with Coefficients Still Attached

Wrong: 2 log(x) + log(y) = log(2xy)

Correct: You CANNOT use the Product Rule if there are numbers in front of the logs. You must use the Power Rule in reverse first to move the coefficient up as an exponent. 2 log(x) + log(y) = log(x2) + log(y) = log(x2y).

Mistake 5: Ignoring Domain Restrictions After Solving

Wrong: Solving log(x) + log(x-3) = 1, getting x = 5 and x = -2, and boxing both answers.

Correct: If you plug x = -2 back into log(x), you get log(-2), which is undefined. You MUST reject x = -2 as an extraneous solution. Only x = 5 is valid.

13. Exam Tips and Problem-Solving Strategies

Strategy 1: Read the Verb Carefully

  • "Evaluate": Find a numerical answer (e.g., log2(8) = 3).
  • "Expand": Tear the log apart. Your final answer should have multiple "log" or "ln" words.
  • "Condense" or "Write as a single logarithm": Squish them together. Your final answer should have exactly one "log" or "ln" word.
  • "Solve": Find the value of x. You will almost certainly need to condense first, then convert to exponential form.

Strategy 2: PEMDAS in Reverse for Condensing

When condensing, always deal with coefficients first (make them exponents). Then deal with addition and subtraction left-to-right to build your numerator and denominator.

Strategy 3: Check Domains BEFORE Solving

Look at the original equation. If you see log(x - 4), immediately write down x > 4 on the side of your paper. Run your final answers against this rule; if an answer isn't greater than 4, cross it out immediately without needing to do a full check.

Strategy 4: Change of Base for the Calculator

If a multiple-choice question asks for the decimal approximation of log5(17), remember the Change of Base formula: type log(17) / log(5) or ln(17) / ln(5) into your calculator. Both will give the exact same correct answer.

14. Practice Problems

Test your understanding. Work through each problem on paper before clicking "Show Solution" to check your method and final answer.

Question 1 (Evaluate)

Evaluate: log4(64)

Think: 4 to what power gives 64?

41 = 4, 42 = 16, 43 = 64.

Answer: 3

Question 2 (Evaluate)

Evaluate: ln(e)

The base of ln is e. So this is asking loge(e).

Using the Log of Base property (logbb = 1).

Answer: 1

Question 3 (Expand)

Expand: log(5x)

Multiplication inside the log. Use the Product Property.

Answer: log(5) + log(x)

Question 4 (Expand)

Expand: log2(x / 8)

Division inside the log. Use the Quotient Property. Then evaluate the numerical part.

= log2(x) - log2(8)

= log2(x) - 3

Answer: log2(x) - 3

Question 5 (Expand)

Expand: ln(x4y2)

Step 1 (Product rule): ln(x4) + ln(y2)

Step 2 (Power rule): Bring exponents to the front.

Answer: 4 ln(x) + 2 ln(y)

Question 6 (Expand)

Expand: log(x3 / (yz))

Step 1 (Quotient rule): log(x3) - log(yz)

Step 2 (Product rule on denominator): log(x3) - (log(y) + log(z))

Step 3 (Distribute negative & Power rule): 3 log(x) - log(y) - log(z)

Answer: 3 log(x) - log(y) - log(z)

Question 7 (Condense)

Condense to a single logarithm: log(a) + log(b)

Addition outside means multiplication inside (Product Property reversed).

Answer: log(ab)

Question 8 (Condense)

Condense: 3 log4(x) - log4(y)

Step 1 (Power rule reversed): Move the 3 up to become an exponent → log4(x3) - log4(y)

Step 2 (Quotient rule reversed): Subtraction outside means division inside.

Answer: log4(x3 / y)

Question 9 (Condense)

Condense: 1/2 ln(x) + ln(y) - 3 ln(z)

Step 1 (Power rule): ln(x1/2) + ln(y) - ln(z3). Note: x1/2 is the square root of x.

Step 2 (Product & Quotient rules): Put positive terms in numerator, negative terms in denominator.

Answer: ln(√x * y / z3) or ln(x1/2y / z3)

Question 10 (Solve)

Solve: log2(x) + log2(4) = 5

Method 1 (Condense): log2(4x) = 5 → 25 = 4x → 32 = 4x → x = 8.

Method 2 (Evaluate knowns): log2(4) = 2. So, log2(x) + 2 = 5 → log2(x) = 3 → 23 = x → x = 8.

Check: log2(8) is valid.

Answer: x = 8

Question 11 (Solve)

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

Condense: log((x+2)(x-1)) = 1 → log(x2 + x - 2) = 1.

Convert: 101 = x2 + x - 2 → 10 = x2 + x - 2 → 0 = x2 + x - 12.

Factor: 0 = (x + 4)(x - 3) → x = -4 or x = 3.

Check Domain: If x = -4, log(-4+2) = log(-2), undefined. Reject. If x = 3, log(5) and log(2), both > 0. Valid.

Answer: x = 3

Question 12 (Domain)

Find the domain: f(x) = ln(2x - 8)

Set argument > 0: 2x - 8 > 0.

Solve: 2x > 8 → x > 4.

Answer: x > 4 (Interval notation: (4, ∞))

Question 13 (Solve)

Solve: ln(x2) = 2 ln(4)

Step 1 (Power rule backwards on right side): ln(x2) = ln(42) → ln(x2) = ln(16)

Step 2 (One-to-one property): Since bases are same, x2 = 16.

Step 3 (Solve): x = √16 = 4 or x = -√16 = -4.

Check: ln((4)2) = ln(16) valid. ln((-4)2) = ln(16) valid.

Answer: x = 4, x = -4

Question 14 (Change of Base)

Evaluate log3(20) to three decimal places.

Use Change of Base formula: log(20) / log(3) OR ln(20) / ln(3).

1.3010 / 0.4771 ≈ 2.7268

Answer: 2.727

Question 15 (Solve)

Solve: log5(3x) - log5(2) = log5(15)

Step 1 (Condense left side): log5(3x / 2) = log5(15)

Step 2 (One-to-one property): 3x / 2 = 15

Step 3 (Solve): 3x = 30 → x = 10.

Check: log5(3*10) = log5(30) valid.

Answer: x = 10

Interactive Logarithmic Properties Helper

Expander & Evaluator

Enter a base (b) and multiple arguments (like M and N) to see how the Product Property expands them and calculates the total.

Results will appear here...

15. Summary of Key Points

Property / Concept Formula Form Explanation
Definition logb(a) = c ↔ bc = a Converts between log and exponential forms.
Product Property logb(MN) = logb(M) + logb(N) Multiply inside = Add outside.
Quotient Property logb(M/N) = logb(M) - logb(N) Divide inside = Subtract outside.
Power Property logb(Mp) = p · logb(M) Exponent on argument moves to front as multiplier.
Log of 1 logb(1) = 0 Any base to power 0 is 1.
Log of Base logb(b) = 1 Any base to power 1 is itself.
Change of Base logb(a) = log(a) / log(b) Used to calculate weird bases using standard calculators.
Domain restrictions Argument > 0 Cannot take log of zero or a negative number.
Extraneous solutions Check answers in original eq Reject answers that force argument ≤ 0.

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