Logarithm rules: the three laws, where each comes from, and how to solve an exponential equation and a logarithmic equation without a false root

There are three logarithm rules, and each is an exponent law read backwards: log(ab) = log a + log b because aˣ · aʸ = aˣ⁺ʸ; log(a/b) = log a − log b because aˣ/aʸ = aˣ⁻ʸ; log(aⁿ) = n log a because (aˣ)ⁿ = aⁿˣ. A logarithm is an exponent, so the rules are the exponent rules. This page derives the three, adds the change-of-base formula, and solves one exponential and one logarithmic equation line by line, including the root you must throw away.

What is a logarithm, in one sentence

A logarithm is the exponent you raise a base to in order to get a number: log₂ 8 = 3 because 2³ = 8.

One instance in each of the three bases you will meet:

• Base 2: log₂ 8 = 3, because 2³ = 8.
• Base 10: log 100 = 2, because 10² = 100. A log with no base written is base 10.
• Base e: ln e = 1, because e¹ = e. The natural log, ln, is the log with base e, about 2.718.

The base must be positive and not 1. A base of 1 gives 1 for every exponent, so it could never reach any other number.

In MHF4U these ideas are 10, "Logarithms as the inverse of exponentiation", 11, "The laws of logarithms", 12, "Solving exponential equations", and 13, "Solving logarithmic equations". They follow idea 9, "Exponential functions and their graphs", and lead into idea 14, "Exponential and logarithmic models". The MHF4U course map shows the whole strand, and the precalculus ideas calculus assumes put logarithms on the list.

The three log rules, and where each comes from

Each rule below starts with an exponent law you already trust, in base 2 so the numbers are easy to check.

The product rule

2³ · 2⁴ = 2⁷. Multiplying powers adds exponents. Logarithms are those exponents, so:

log₂(8 · 16) = log₂ 8 + log₂ 16 = 3 + 4 = 7. And 8 · 16 = 128 = 2⁷. It checks.

The law: log(ab) = log a + log b.

The quotient rule

2⁷ ÷ 2³ = 2⁴. Dividing powers subtracts exponents. So:

log₂(128/8) = log₂ 128 − log₂ 8 = 7 − 3 = 4. And 128/8 = 16 = 2⁴.

The law: log(a/b) = log a − log b.

The power rule

(2³)² = 2⁶. A power of a power multiplies exponents. So:

log₂(8²) = 2 · log₂ 8 = 2 · 3 = 6. And 8² = 64 = 2⁶.

The law: log(aⁿ) = n log a. This is the rule that solves exponential equations, because it brings an exponent down to the ground.

The two rules students invent

These look like log laws and are not. Each fails on numbers.

• log(a + b) is not log a + log b. log₂(8 + 8) = log₂ 16 = 4, but log₂ 8 + log₂ 8 = 3 + 3 = 6.
• log a · log b is not log(ab). log₂ 8 · log₂ 16 = 3 · 4 = 12, but log₂(8 · 16) = log₂ 128 = 7.

A product inside becomes a sum outside. Nothing turns a sum inside into anything simpler.

Change of base

Your calculator has two logs, ln and log (base 10). For any other base, use change of base: the log of x in base b equals ln x ÷ ln b.

Why it works: if y is the base-b log of x, then x = bʸ. Take ln of both sides: ln x = y ln b by the power rule. Divide: y = ln x ÷ ln b.

One instance, which the next section needs: log₃ 8 = ln 8 ÷ ln 3 ≈ 2.0794 ÷ 1.0986 ≈ 1.893.

Worked example 1: how to solve exponential equations

Solve 5 · 3ˣ = 40.

1. Isolate the power. Divide both sides by 5: 3ˣ = 8.
2. Take a logarithm of both sides. Any base works; use ln: ln(3ˣ) = ln 8.
3. Use the power rule to bring x down. x ln 3 = ln 8.
4. Divide. x = ln 8 ÷ ln 3, which you can also write 3 ln 2 ÷ ln 3.
5. Evaluate. x ≈ 2.07944 ÷ 1.09861 ≈ 1.893.

Check: 3 raised to the power 1.893 is about 8, and 5 × 8 = 40.

The common wrong line, first version: taking the log before isolating, and writing ln(5 · 3ˣ) = x ln 15. That treats 5 · 3ˣ as 15ˣ, which it is not. Following it gives x = ln 40 ÷ ln 15 ≈ 1.362, a different and wrong answer. The product rule gives ln 5 + x ln 3, which does work, but isolating first is cleaner and harder to get wrong.

The common wrong line, second version: x = 8/3, because "the 3 and the x are together". The x is an exponent. Division does not reach an exponent; only a logarithm brings it down.

Worked example 2: how to solve logarithmic equations

Solve log₂(x) + log₂(x − 2) = 3.

1. Combine the left side with the product rule. log₂(x(x − 2)) = 3.
2. Rewrite in exponential form. x(x − 2) = 2³ = 8.
3. Expand and set to zero. x² − 2x − 8 = 0.
4. Factor. (x − 4)(x + 2) = 0, so x = 4 or x = −2.
5. Check each root in the original. x = 4: log₂ 4 + log₂ 2 = 2 + 1 = 3. It works. x = −2: log₂(−2) is undefined. Reject it.
6. State the answer. x = 4.

The common wrong line, first version: keeping x = −2. The algebra produced it, but the original equation has no meaning there, because a logarithm of a negative number or of zero is undefined. Merging the two logs in line 1 created the extra root. Booth and colleagues (2013) found that seeing an incorrect example with its fault explained may help understanding more than seeing only correct ones; this is the incorrect example that costs the most marks.

The common wrong line, second version: writing log₂ x + log₂(x − 2) = log₂(2x − 2), adding the arguments. The product rule multiplies them.

Siblings to try cold

Cover both examples. Write every line, then check. A worked example is fastest when it is followed by your own attempt, which is what Sweller and Cooper (1985) found for beginners in algebra.

1. Exponential: solve 2 · 5ˣ = 70.
2. Logarithmic: solve log₃(x) + log₃(x + 6) = 3.

Check

1. 5ˣ = 35, so x = ln 35 ÷ ln 5 ≈ 2.209.
2. x(x + 6) = 27 gives (x + 9)(x − 3) = 0. x = 3 checks: log₃ 3 + log₃ 9 = 1 + 2 = 3. x = −9 is rejected. The answer is x = 3.

If you kept −9, go back to the check line. That line is the mark.

Why the log of a negative number is undefined

For a positive base, bˣ is positive for every real x. A large negative exponent gives a tiny positive number, never zero and never negative. So no exponent produces −2 or 0, and the logarithm, which asks for that exponent, has no answer.

That is why the domain of a logarithm is x > 0, and why every logarithmic equation ends with a check. It is the same habit rational equations need: reject any root the original equation cannot accept.

Where it goes next

Idea 14, exponential and logarithmic models, uses example 1's method on half-lives and doubling times. Logarithms and exponentials undo each other, which is the swap described in domain and range, inverses and transformations. In calculus, the derivative of ln x is 1/x, and the integral of 1/x brings the log back.

Learn the three laws by deriving them on a blank page, not by rereading them; testing yourself before you reread explains why. Then work a few of each equation type, as worked examples, then practice problems lays out. If the false root keeps getting you, chapter 4 of the free playbook is about finding the line. Other MHF4U ideas are on the math help hub.

Log rules on one screen

• A logarithm is an exponent: log₂ 8 = 3 because 2³ = 8.
• Product: log(ab) = log a + log b, from aˣ · aʸ = aˣ⁺ʸ.
• Quotient: log(a/b) = log a − log b, from aˣ/aʸ = aˣ⁻ʸ.
• Power: log(aⁿ) = n log a, from (aˣ)ⁿ = aⁿˣ.
• Not laws: log(a + b) ≠ log a + log b; log a · log b ≠ log(ab).
• Change of base: base-b log of x = ln x ÷ ln b.
• Exponential equations: isolate the power, then log both sides.
• Logarithmic equations: merge, exponentiate, solve, then check every root.