Logarithmic Functions

A logarithm is not a new kind of monster. It is an exponent wearing a different hat. Every time you ask "two raised to what power gives me thirty-two?" you are asking a logarithm question — you have been doing this since you first learned exponents, just without the vocabulary. This chapter gives you the vocabulary, shows you how to move a statement back and forth between exponential form and logarithmic form, teaches you to read a logarithmic graph, and hands you three properties that let you take a complicated logarithm apart or squeeze several logarithms into one. Go slowly, say the definition out loud a few times, and the rest follows.

Section 1What a logarithm is

Undoing an exponential

Start with a question you already know how to answer: 2 raised to what power gives 8? You said 3 instantly, because 2³ = 8. You just evaluated a logarithm. The only thing missing was a symbol for "the power you raise 2 to," and that symbol is log.

Here is the whole definition, and it is worth memorizing word for word:

log_b(x) = y means exactly the same thing as b^y = x.

Read the left side out loud as "log base b of x equals y." Read what it means as "y is the exponent you put on b to get x." If you can only remember one sentence from this chapter, remember this one: a logarithm is an exponent. The answer to any log question is always an exponent, and that is why logs behave the way they do for the rest of the chapter.

Why do we need this at all? Because an exponential function like f(x) = 2^x is one-to-one — every input gives a different output — so it has an inverse function, and that inverse is the logarithm. Exponentials answer "I have the exponent, what is the result?" Logarithms answer "I have the result, what was the exponent?" Each undoes the other:

log_b(b^x) = x   and   b^log_b(x) = x

Converting between the two forms

Almost every logarithm problem in this chapter starts with a conversion, so make it automatic. The trick that works for most students: the base stays the base, and the logarithm's answer becomes the exponent. Draw a little loop from the base, up over the equals sign, to the answer.

The rules about what is allowed

Two restrictions come with the definition, and neither is arbitrary.

First, the base must satisfy b > 0 and b ≠ 1. A negative base breaks down as soon as you use fractional exponents (what would (−4)^1/2 be?), and a base of 1 is useless because 1^y = 1 for every y — the question "1 to what power gives 8?" has no answer at all.

Second, the argument — the thing inside the log — must satisfy x > 0. This one trips people up until they see why. Since b is positive, b^y is always positive, no matter what y is. A positive number raised to any power — positive, negative, zero, fractional — can never land on zero or on a negative number. So there is no exponent that makes 2^y equal −8, and log₂(−8) is simply undefined. Keep that in your pocket; in Section 3 it becomes the rule that determines the domain of every logarithmic function.

A fully worked example

Solve log₅(2x − 3) = 2.

Step 1 — Convert to exponential form. The base is 5, the answer is 2, so 2 becomes the exponent on 5:

log₅(2x − 3) = 2   becomes   5² = 2x − 3

Step 2 — Do the arithmetic on the left. 5² = 25, so

25 = 2x − 3

Step 3 — Add 3 to both sides.

25 + 3 = 2x − 3 + 3  →  28 = 2x

Step 4 — Divide both sides by 2.

28 / 2 = 2x / 2  →  14 = x

Step 5 — Check it, including the domain. Put x = 14 back into the original argument: 2(14) − 3 = 28 − 3 = 25. Is 25 > 0? Yes, so the argument is legal. And log₅(25) = 2, because 5² = 25. The left side equals the right side, so x = 14 is the solution.

Section 2Evaluating logarithms

Two bases get their own shorthand

Any positive base (other than 1) is allowed, but two bases show up so often that mathematicians stopped writing them.

The common logarithm has base 10, and we write it with no base at all: log(x) means log₁₀(x). It shows up wherever we measure things in powers of ten — decibels, pH, the Richter scale, orders of magnitude.

The natural logarithm has base e, where e ≈ 2.718281828. We write it ln(x), which means log_e(x). The number e looks strange the first time you meet it, but it is just a fixed number like π — irrational, non-repeating, and enormously useful because it is the natural base of continuously compounding processes — the factor a quantity growing continuously at 100% per unit time multiplies by in one unit of time. On your calculator, LOG is base 10 and LN is base e. They are different buttons and they give different answers; mixing them up is the single most common calculator error in this chapter.

Evaluating by inspection

Most logarithms you meet in a homework set are meant to be done in your head. The method never changes: set the log equal to y, convert to exponential form, and rewrite both sides with the same base.

Four shortcuts fall straight out of the definition and are worth knowing cold:

Two more patterns are worth practicing because they look harder than they are. A reciprocal argument gives a negative answer: log₂(1/8) = −3, because 2⁻³ = 1/(2³) = 1/8. And a root gives a fractional answer: log₉(3) = ½, because 9^1/2 = √9 = 3.

A fully worked example

Evaluate log₈(4) without a calculator.

This one has no obvious answer — 8 to the first power is too big, 8 to the zero power is too small — so the answer must be a fraction between 0 and 1. Here is the full method.

Step 1 — Name the answer. Let y = log₈(4).

Step 2 — Convert to exponential form.

8^y = 4

Step 3 — Rewrite both sides with a common base. Both 8 and 4 are powers of 2: 8 = 2³ and 4 = 2². So

(2³)^y = 2²

Step 4 — Use the power-of-a-power rule (multiply the exponents):

2^3y = 2²

Step 5 — Match the exponents. If the bases are equal and the expressions are equal, the exponents must be equal:

3y = 2

Step 6 — Solve. Divide both sides by 3:

y = (2)/(3)

Step 7 — Check. Does 8^2/3 really equal 4? A fractional exponent means "root on the bottom, power on the top": 8^2/3 = (³√8)² = (2)² = 4. It checks. So log₈(4) = 2/3, and notice it landed between 0 and 1 exactly as we predicted.

Section 3Graphs of logarithmic functions

The shape of the parent function

Graph f(x) = log₂(x) by picking convenient x-values — convenient here means powers of 2, so the logs come out clean.

Plot those and a clear picture appears. The curve rises forever but rises more and more slowly — to gain one more unit of height you have to double your x. On the left side, as x shrinks toward 0, the curve plunges downward without limit, hugging the y-axis but never touching it. That vertical line x = 0 is the vertical asymptote.

Every parent logarithmic function y = log_b(x) with b > 1 shares these features:

• Domain: x > 0, written (0, ∞) — you can never feed it zero or a negative number.
• Range: all real numbers, (−∞, ∞) — the output can be anything.
• Vertical asymptote: the line x = 0.
• x-intercept: the point (1, 0), because log_b(1) = 0 for every base.
• Anchor point: the point (b, 1), because log_b(b) = 1.
• No y-intercept — x = 0 is not in the domain of the parent function. (Shift the graph horizontally and that can change; see the worked example below.)
• The function is increasing when b > 1 and decreasing when 0 < b < 1.

Compare that list to the exponential function y = b^x, whose domain is all reals, whose range is y > 0, and whose asymptote is the horizontal line y = 0. Every feature is swapped, because the two graphs are mirror images across the line y = x. That is what "inverse" looks like on paper.

Transformations, and where the asymptote goes

The general form is y = a·log_b(x − h) + k. Each letter does one job, and the one you must never lose track of is h, because h drags the asymptote with it.

Finding the domain is always the same one-line job: set the argument greater than zero and solve. For f(x) = log(3x − 12), require 3x − 12 > 0, so 3x > 12, so x > 4. Domain: (4, ∞), and the vertical asymptote is x = 4.

One warning before you do this on autopilot. When the variable carries a negative coefficient, solving the domain inequality makes you divide by a negative number, and dividing (or multiplying) both sides of an inequality by a negative number reverses the inequality sign. For f(x) = log(8 − 2x), require 8 − 2x > 0, so −2x > −8, and dividing by −2 flips the sign: x < 4. Domain: (−∞, 4), with the asymptote still at x = 4 — but now the curve lives to the left of it. Test a point to be sure: x = 0 gives log(8), which is defined, and 0 is indeed less than 4. ✓

A fully worked example

For f(x) = log₂(x + 3) − 1, find the domain, the vertical asymptote, the x-intercept, and two more plotting points.

Step 1 — Domain. The argument must be positive:

x + 3 > 0  →  x > −3.   Domain: (−3, ∞).

Step 2 — Vertical asymptote. The asymptote sits where the argument equals zero: x + 3 = 0, so x = −3. (Notice the parent asymptote x = 0 moved left 3, matching the "+3" inside.)

Step 3 — x-intercept. Set f(x) = 0 and solve:

log₂(x + 3) − 1 = 0

Add 1 to both sides:   log₂(x + 3) = 1

Convert to exponential form:   x + 3 = 2¹ = 2

Subtract 3 from both sides:   x = 2 − 3 = −1

So the x-intercept is (−1, 0). Check: f(−1) = log₂(−1 + 3) − 1 = log₂(2) − 1 = 1 − 1 = 0. ✓

Step 4 — Two more points. Choose x-values that make x + 3 a power of 2.

Let x = −2:   f(−2) = log₂(−2 + 3) − 1 = log₂(1) − 1 = 0 − 1 = −1. Point: (−2, −1).

Let x = 1:   f(1) = log₂(1 + 3) − 1 = log₂(4) − 1 = 2 − 1 = 1. Point: (1, 1).

Step 5 — Assemble. Draw the dashed vertical line x = −3, plot (−2, −1), (−1, 0), and (1, 1), and sketch a curve that dives down along the asymptote on the left and rises slowly to the right. The range is still all real numbers — a vertical shift never changes that.

One bonus feature worth noticing: because the shift moved the asymptote to x = −3, the value x = 0 is now inside the domain, so this graph does have a y-intercept. Evaluate it: f(0) = log₂(0 + 3) − 1 = log₂(3) − 1 ≈ 1.585 − 1 = 0.585, giving roughly (0, 0.585). The parent graph had no y-intercept, but a horizontal shift can hand you one — so check the domain rather than reciting a rule.

Section 4Properties of logarithms

Three rules, inherited from exponents

Because a logarithm is an exponent, every exponent rule you already know has a logarithm twin. There are three, and all of them assume M > 0, N > 0, b > 0, and b ≠ 1.

Here is where the product rule comes from, in four lines. Let m = log_b(M) and n = log_b(N). Converting to exponential form, M = b^m and N = bⁿ. Multiply them: MN = b^m·bⁿ = b^m+n. Convert that back to logarithmic form: log_b(MN) = m + n = log_b(M) + log_b(N). The quotient and power rules are proved the same way, from the matching exponent rule.

Notice the pattern in plain English: a product inside becomes a sum outside, a quotient inside becomes a difference outside, and an exponent inside becomes a multiplier out front. That single sentence is the engine of Section 5.

Three ways to break the rules

These three errors account for most lost points on logarithm tests. Read them once carefully now and you will recognize them later.

Error 1: log_b(M + N) is not log_b(M) + log_b(N). The product rule is about multiplication inside, not addition. Test it: log(5) + log(5) = log(25) by the product rule, but log(5 + 5) = log(10) = 1, and log(25) ≈ 1.3979. Those are different numbers, so the two expressions are genuinely not equal. A sum inside a logarithm cannot be broken apart at all.

Error 2: (log_b(M))/(log_b(N)) is not log_b(M/N), and it is not log_b(M) − log_b(N) either. The quotient rule needs the division to happen inside one logarithm; here the division happens outside, between two finished logarithms. Check with base 2, M = 8, N = 2: the quotient is (log₂(8))/(log₂(2)) = 3/1 = 3, while log₂(8/2) = log₂(4) = 2. Not equal. Dividing two separate logarithms is a different operation entirely — in fact that quotient is the change-of-base formula from Section 5, and it equals log_N(M).

Error 3: (log_b(M))² is not 2·log_b(M). The power rule only applies when the exponent is inside, on the argument. log_b(M²) = 2·log_b(M), but squaring the whole logarithm is something else. Check with M = 100, base 10: log(100) = 2, so (log(100))² = 4, while log(100²) = log(10000) = 4 as well — here they happen to match, which is exactly why you cannot trust one example. Try M = 1000: (log(1000))² = 3² = 9, but log(1000²) = log(1000000) = 6. Not equal.

A fully worked example

Given log(2) ≈ 0.3010 and log(3) ≈ 0.4771, find log(72) without a calculator.

Step 1 — Factor 72 into the numbers you were given.

72 = 8 · 9 = 2³ · 3²

Step 2 — Rewrite the logarithm.

log(72) = log(2³ · 3²)

Step 3 — Apply the product rule (multiplication inside becomes addition outside):

log(2³ · 3²) = log(2³) + log(3²)

Step 4 — Apply the power rule twice (exponents come out front):

= 3·log(2) + 2·log(3)

Step 5 — Substitute the given values.

= 3(0.3010) + 2(0.4771)

Step 6 — Multiply.

3(0.3010) = 0.9030   and   2(0.4771) = 0.9542

Step 7 — Add.

0.9030 + 0.9542 = 1.8572

So log(72) ≈ 1.8572.

Step 8 — Check that it is reasonable. Since 10¹ = 10 and 10² = 100, and 72 sits between 10 and 100, the answer must sit between 1 and 2. It does. (A calculator gives 1.85733, so we are right to three decimal places — the small gap in the fourth is only the rounding in the values we were handed.)

Section 5Expanding and condensing

Expanding: taking one logarithm apart

Expanding means rewriting a single complicated logarithm as a sum and difference of simpler ones, with no products, quotients, or exponents left inside. Work outside-in and follow this order every time:

1. Apply the quotient rule first, to split the big fraction into a difference.
2. Apply the product rule to split any remaining multiplications into sums.
3. Apply the power rule to bring every exponent (including roots written as fractional exponents) out front.
4. Evaluate any logarithm that comes out to a plain number.

Fully worked: expand log₃((9x⁴)/(√y)), where x > 0 and y > 0.

Step 1 — Quotient rule. One fraction inside becomes a subtraction outside:

log₃((9x⁴)/(√y)) = log₃(9x⁴) − log₃(√y)

Step 2 — Product rule on the first piece, since 9x⁴ is 9 times x⁴:

= log₃(9) + log₃(x⁴) − log₃(√y)

Step 3 — Rewrite the radical as a fractional exponent, because the power rule cannot see a radical sign:

= log₃(9) + log₃(x⁴) − log₃(y^1/2)

Step 4 — Power rule on both variable terms:

= log₃(9) + 4·log₃(x) − ½·log₃(y)

Step 5 — Evaluate what you can. log₃(9) = 2, because 3² = 9. Final answer:

2 + 4·log₃(x) − ½·log₃(y)

Watch the subtraction sign in Step 1. If the denominator had contained a product — say √y·z — then both resulting pieces would be subtracted, because the whole denominator is what gets taken away.

Condensing: squeezing several logarithms into one

Condensing runs the same machinery backward. The goal is a single logarithm with coefficient 1. The order reverses:

1. Use the power rule first, moving every coefficient up into an exponent. Nothing else can proceed until every log has coefficient 1.
2. Use the product rule to combine every term being added into one logarithm of a product.
3. Use the quotient rule to move every term being subtracted into the denominator.

Fully worked: condense 3·log₂(x) + ½·log₂(y) − 2·log₂(z).

Step 1 — Power rule on all three terms. Each coefficient becomes an exponent:

= log₂(x³) + log₂(y^1/2) − log₂(z²)

Step 2 — Product rule on the two added terms. Addition outside becomes multiplication inside:

= log₂(x³ · y^1/2) − log₂(z²)

Step 3 — Quotient rule. Subtraction outside becomes division inside:

= log₂((x³ · y^1/2)/(z²))

Step 4 — Write the fractional exponent as a radical if you prefer:

= log₂((x³√y)/(z²))

Step 5 — Check by expanding it back. Quotient rule gives log₂(x³y^1/2) − log₂(z²); product rule gives log₂(x³) + log₂(y^1/2) − log₂(z²); power rule gives 3·log₂(x) + ½·log₂(y) − 2·log₂(z). Back where we started. ✓

Change of base

Your calculator has only two logarithm buttons, base 10 and base e. So what do you do with log₅(89)? You change the base:

log_b(M) = (log_c(M))/(log_c(b)) for any legal new base c

This is a fraction, so it is fair to ask when the denominator could be zero. It never can: log_c(b) = 0 would force b = 1, and b = 1 was already ruled out as a base back in Section 1. Both bases also have to stay positive, and M > 0 as always. So the formula is defined for every logarithm you are allowed to write in the first place.

In practice c is 10 or e, giving the two forms you will actually type: log_b(M) = (log(M))/(log(b)) or log_b(M) = (ln(M))/(ln(b)). Both give the same number. The memory hook: the old base goes on the bottom.

Fully worked: evaluate log₅(89) to four decimal places.

Step 1 — Estimate first. 5² = 25 and 5³ = 125. Since 89 sits between 25 and 125, the answer must sit between 2 and 3.

Step 2 — Apply change of base with natural logs (old base 5 goes on the bottom):

log₅(89) = (ln(89))/(ln(5))

Step 3 — Evaluate each piece.

ln(89) ≈ 4.48864   and   ln(5) ≈ 1.60944

Step 4 — Divide.

4.48864 / 1.60944 ≈ 2.78895

Step 5 — Round and check. log₅(89) ≈ 2.7889, which lands between 2 and 3 exactly as predicted. Verify by reversing it: 5^2.7889 ≈ 89. ✓

One more use: solving equations, and checking for extraneous solutions

Condensing is the standard first move when an equation has two logarithms on one side. But logarithms have a domain restriction, so the algebra can hand you an answer the original equation cannot accept. Those are extraneous solutions, and you must check for them every single time.

Fully worked: solve log₅(x) + log₅(x − 4) = 1.

Step 1 — Note the domain up front. We need x > 0 and x − 4 > 0, so x > 4. Any answer at or below 4 will have to be thrown out.

Step 2 — Condense the left side with the product rule:

log₅(x(x − 4)) = 1

Step 3 — Convert to exponential form.

x(x − 4) = 5¹ = 5

Step 4 — Distribute and set equal to zero.

x² − 4x = 5  →  x² − 4x − 5 = 0

Step 5 — Factor. We need two numbers that multiply to −5 and add to −4: those are −5 and +1.

(x − 5)(x + 1) = 0  →  x = 5 or x = −1

Step 6 — Test both against the domain. x = −1 fails, because log₅(−1) is undefined — it is extraneous and gets discarded. x = 5 passes, since 5 > 4.

Step 7 — Verify the survivor in the original equation.

log₅(5) + log₅(5 − 4) = log₅(5) + log₅(1) = 1 + 0 = 1 ✓

The solution is x = 5, and only x = 5.

ReferenceWords worth knowing

A quick reference for the terms in this chapter — skim it now, or circle back whenever a word trips you up.

Argument

The quantity inside the logarithm. It must always be strictly greater than zero.

Asymptote, vertical

A vertical line the graph approaches without ever reaching. For y = log_b(x − h) it is the line x = h.

Base

The number being raised to a power. In log_b(x) it is b, and it must satisfy b > 0 and b ≠ 1.

Change-of-base formula

log_b(M) = (log_c(M))/(log_c(b)); it lets a calculator evaluate a logarithm of any base.

Common logarithm

A logarithm with base 10, written log(x) with the base left off.

Condensing

Combining a sum and difference of logarithms into one logarithm with coefficient 1.

Domain

The set of allowed inputs. For a logarithmic function it is every x that makes the argument positive.

e

The irrational constant approximately 2.718281828, the base of the natural logarithm.

Expanding

Rewriting one logarithm as a sum and difference of simpler logarithms with no products, quotients, or exponents inside.

Exponential form

The statement b^y = x, equivalent to the logarithmic form log_b(x) = y.

Extraneous solution

A value produced by correct algebra that fails in the original equation, usually because it makes a logarithm's argument zero or negative.

Inverse function

A function that undoes another. The logarithm and the exponential with the same base are inverses, and their graphs are mirror images across y = x.

Logarithm

The exponent to which a base must be raised to produce a given number: log_b(x) = y means b^y = x.

Logarithmic form

The statement log_b(x) = y, equivalent to the exponential form b^y = x.

Natural logarithm

A logarithm with base e, written ln(x).

One-to-one function

A function in which no two different inputs share an output. Exponential functions are one-to-one, which is why they have inverses.

Power rule

log_b(M^p) = p·log_b(M) — an exponent inside becomes a multiplier outside.

Product rule

log_b(MN) = log_b(M) + log_b(N) — multiplication inside becomes addition outside.

Quotient rule

log_b(M/N) = log_b(M) − log_b(N) — division inside becomes subtraction outside.

Range

The set of possible outputs. For every parent logarithmic function it is all real numbers.

x-intercept

Where a graph crosses the x-axis. For y = log_b(x) it is always (1, 0).