1
A logarithm is an exponent
Not a length, not an operation on digits, not a button. log_5(125) is the answer to "5 to what power gives 125?", and the answer is 3. Read every logarithm you meet as that question.
Lesson · Exponential and Logarithmic Functions · Logarithms
What a Logarithm Is
log_b(x) is the exponent you put on b to get x. Everything later in this course — the graphs, the three laws, both kinds of equation — is that one sentence applied, and nearly every error is a step taken without it.
log_b(x) = y means exactly b^y = x
with b > 0, b ≠ 1, x > 0
log_2(8) = 3 because 2³ = 8
log_9(3) = 1/2 because 9^(1/2) = 3
One fact, written two ways
Core ideas
What this lesson establishes
Say the definition out loud until it is automatic. Everything in this lesson is a consequence of it, including the two restrictions.
2
The two forms carry identical information
b^y = x and log_b(x) = y are the same statement. Moving between them is not a step in a calculation, it is a change of notation, and it costs nothing.
3
The argument must be positive
b^y is positive for every real y, so no exponent produces 0 and none produces a negative number. log_2(0) and log_2(−8) name nothing — not because they are difficult, but because the question they ask has no answer.
The material
The definition, read in both directions
One definition, the restrictions that come with it, and the evaluations you should be able to do without a calculator.
Definition · Logarithm
For b > 0 with b ≠ 1, and for x > 0, log_b(x) is the unique real number y with b^y = x. The number b is the base of the logarithm and x is its argument.
The word unique is carrying weight. “Exponential Functions” fenced the base in so that b^y is strictly increasing (when b > 1) or strictly decreasing (when 0 < b < 1). Either way it takes each positive value exactly once, so there is one y and not two. That is also why b = 1 is barred: 1^y = 1 for every y, so log_1(1) would have every number as an answer and log_1(5) would have none.
exponential form logarithmic form
────────────────────────────────────────────────
b^y = x log_b(x) = y
2³ = 8 log_2(8) = 3
10^(−3) = 1/1000 log_10(1/1000) = −3
5^0 = 1 log_5(1) = 0
7^1 = 7 log_7(7) = 1
9^(1/2) = 3 log_9(3) = 1/2
Read the table left to right and you are converting a power into a logarithm; read it right to left and you are doing the reverse. Three roles move around: the base of the power stays the base of the logarithm, the result of the power becomes the argument, and the exponent becomes the value. Nothing else happens.
Theorem · The two cancellation identities
b^(log_b(x)) = x for every x > 0, and log_b(b^y) = y for every real y. The first says that raising b to the exponent which produces x produces x. The second says that the exponent which produces b^y is y. Both are the definition said aloud, and “Logarithmic Functions and Their Graphs” restates them as the statement that the two functions are inverse.
Evaluating without a calculator
Ask the question. log_3(81): three to what power is 81? 3⁴ = 81, so the value is 4. When the answer is not a whole number the same question still works, provided both numbers can be written as powers of one common base.
Example · log_8(4), in full
Both 8 and 4 are powers of 2: 8 = 2³ and 4 = 2². If 8^y = 4 then (2³)^y = 2², so 2^(3y) = 2². Since 2^t takes each value exactly once, 3y = 2 and y = 2/3. Check: 8^(2/3) = (∛8)² = 2² = 4, so log_8(4) = 2/3.
With a base greater than 1, a logarithm is negative exactly when the argument lies strictly between 0 and 1; with a base below 1 it is negative exactly when the argument is above 1. log_5(1/25) = −2, because 5^(−2) = 1/25. The restriction is on the argument; the value is free to be any real number at all.
Most logarithms are irrational. log_2(10) sits between 3 and 4, because 2³ = 8 and 2⁴ = 16 bracket 10, and its value is 3.321928… — a number with no exact decimal form. The exact statement is 2^(log_2(10)) = 10; the decimal is a rounding, and “Change of Base” is about producing it.
Notation. The base is written as a subscript, and the argument gets brackets whenever there is any doubt: log_2(x + 1) and log_2(x) + 1 are different numbers. Two bases are used so often that they drop the subscript — base 10 and base e — and “Common and Natural Logarithms” introduces that shorthand.
Interactive
Two ways to say one thing
log_b(x) = y and b^y = x are the same sentence
Method
Evaluating a logarithm by hand
Four steps, and the last one is the one that catches a flipped fraction.
1
Write the question as an equation
log_b(x) asks "b to what power is x?". Put it in exponential form, b^y = x, with y the unknown. Everything after this is ordinary algebra.
2
Write both sides as powers of one base
log_8(4) becomes 2^(3y) = 2². Whenever b and x are powers of a common number this works, and the answer comes out as a fraction.
3
Match the exponents
b^m = b^n forces m = n when b > 0 and b ≠ 1, because such an exponential takes each value exactly once. Solve the linear equation that leaves.
4
Check by raising the base to your answer
8^(2/3) = 4. One line, and it catches the inverted fraction, which is the usual way this goes wrong.
Reading the definition
Score 0 / 0Choose the answer you can justify.
Eight logarithms, by asking the question
No calculator is used or needed. The middle column is the question; the right column is the answer to it.
log_2(8) 2 to what power is 8? 2³ = 8 3
log_3(81) 3 to what power is 81? 3⁴ = 81 4
log_5(1/25) 5 to what power is 1/25? 5^(−2) = 1/25 −2
log_9(3) 9 to what power is 3? 9^(1/2) = 3 1/2
log_8(4) 8 to what power is 4? 8^(2/3) = 4 2/3
log_(1/2)(8) 1/2 to what power is 8? (1/2)^(−3) = 8 −3
log_6(1) 6 to what power is 1? 6^0 = 1 0
log_7(−7) 7 to what power is −7? nothing works undefined
checks: 9^(1/2) = √9 = 3 ✓
8^(2/3) = (∛8)² = 2² = 4 ✓
(1/2)^(−3) = 2³ = 8 ✓
The sixth line is the one that surprises people. A base below 1 gives a negative logarithm for an argument above 1, and a positive one for an argument below 1 — the whole scale runs backwards, which is exactly what “Logarithmic Functions and Their Graphs” sees as a decreasing curve.
The last line is not a hard case, it is an impossible one. 7^y is positive for every real y, so no exponent gives −7. Logarithms of negative numbers are not covered in this course and are not merely difficult here; within the real numbers there is nothing to compute.
For a faded evaluation, start log_27(9) = y from the supplied rewrite 27^y = 9, then express both sides as powers of 3, solve for y, and check by raising 27 to the result. Keep the answer as an exact fraction and say which role each of 27, 9 and y plays.
Common mistakes
1
Swapping the base and the argumentlog_2(8) = 3 and log_8(2) = 1/3. They are reciprocals, not equals. Say the base out loud when you read the symbol: it is the number being raised to a power.
2
Expecting log_b(0) to be 0, or −∞No real exponent makes 2^y equal 0; the values shrink without ever arriving. log_2(0) is undefined, and −∞ is not a real number that can serve as an answer.
3
Refusing a negative logarithmlog_5(1/25) = −2 is an ordinary value. The positivity restriction is on the argument. The output of a logarithm runs over every real number, negatives included, which is the range statement of “Logarithmic Functions and Their Graphs”.
Check your understanding
✓
Finish when you convert between the two forms without thinking.
Given log_4(x) = 3/2 you should write 4^(3/2) = x and get 8 in one line. Given 3^y = 1/27 you should write y = log_3(1/27) = −3 in one. If either takes a moment of translation, the definition is not yet automatic and the laws in “The Laws of Logarithms” will feel arbitrary.
Note: Every exact logarithm the lab reports was found by searching for a rational exponent, never by calling a logarithm function. That is the honest shape of the subject: log_2(8) is 3 exactly, and log_2(10) is a number that can only be bracketed and then rounded.