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Exponential and Logarithmic Functions

Quantities that grow or shrink by the same percent each step, the logarithms that undo them, and how to solve equations where the unknown sits in an exponent.

An exponential function like y = 2^x multiplies by the same number every time x goes up by 1. A logarithm answers the reverse question: what power gives this number? Since 2³ = 8, we write log₂ 8 = 3. Logs are the inverse of exponentials, so their graphs are mirror images across the line y = x.

Start: growth and decay Jump to logarithms Jump to practice

Teal: y = 2^x passes through (0, 1), (1, 2) and (2, 4). Plum: y = log₂ x passes through (1, 0), (2, 1) and (4, 2). Swapping x and y turns one graph into the other, a reflection across the dashed line y = x.

Growth and decay

An exponential model has the form y = a · b^x. The number a is the starting value (the value of y when x = 0) and b is the growth factor, what you multiply by each step. If b > 1 the model shows growth; if b is between 0 and 1 it shows decay. A percent rate r, written as a decimal, gives the factor: b = 1 + r for growth and b = 1 − r for decay.

Teal: y = 2^x doubles each step, so it grows. Coral: y = (12)^x halves each step, so it decays. Both pass through (0, 1).

Example 1: Growth or decay, and at what rate?

y = 300 · 1.06^xb = 1.06 > 1, so growth: 1.06 = 1 + 0.06, a 6% ratey = 50 · 0.92^xb = 0.92 < 1, so decay: 0.92 = 1 − 0.08, an 8% rate

Example 2: A car that loses 15% of its value a year

y = 24000 · 0.85^tIt starts at $24,000 and keeps 85% each yeary = 24000 · 0.85⁴After 4 yearsy ≈ 12528.15About $12,528

Compound interest and the number e

If P dollars earn a yearly rate r compounded n times a year, after t years the amount is A = P(1 + r/n)^nt. Compounding more often helps a little, but not without limit. The values of (1 + 1/n)ⁿ settle down toward a number near 2.718, called e. When interest is compounded continuously, A = P · e^rt.

Values of (1 + 1/n)^n
n	(1 + 1/n)^n
1	2
12	2.61304 (rounded)
365	2.71457 (rounded)
1000000	2.71828 (rounded)

Example 3: $2000 at 4% compounded monthly for 5 years

A = 2000(1 + 0.0412)^{12 · 5}P = 2000, r = 0.04, n = 12, t = 5A ≈ 2441.99About $2,441.99. Compounded yearly it would be about $2,433.31

Example 4: The same money compounded continuously

A = 2000 · e^{0.04 · 5}Use A = P · e^rtA = 2000 · e^0.2A ≈ 2442.81Only about 82 cents more than monthly

Logarithms as inverses

For a base b with b > 0 and b ≠ 1, log_b x = y means exactly the same as b^y = x. The log is the exponent. Because the two undo each other, log_b(b^x) = x and b^log_b x = x. You can only take the log of a positive number. Two bases have special names: the common log log x means base 10, and the natural log ln x means base e.

Example 5: Write in log form

2⁵ = 32The base is 2 and the exponent is 5log₂ 32 = 5The log equals the exponent

Example 6: Write in exponential form

log₃ 81 = 4The base is 3 and the log is 43⁴ = 81

Example 7: Evaluate logs

log₅(125) = −25⁻² = 125log 1000 = 310³ = 1000ln(e⁴) = 4ln undoes e to a power

Log rules

Logs turn multiplying into adding, because exponents add when you multiply powers. For positive M and N:

Product rule: log_b(MN) = log_b M + log_b N
Quotient rule: log_b(M/N) = log_b M − log_b N
Power rule: log_b(M^k) = k · log_b M
Change of base: log_b M = log M/log b, which lets a calculator find any log

Example 8: Expand a log

log₂(8x³) = log₂ 8 + log₂(x³)Product rule, for x > 0log₂(8x³) = 3 + 3log₂ xlog₂ 8 = 3, and the power rule brings the 3 down

Example 9: Condense and evaluate

log₃ 54 − log₃ 2 = log₃(542)Quotient rulelog₃ 27 = 33³ = 27

Example 10: Use change of base

log₅ 30 = log 30log 5Change of baselog₅ 30 ≈ 2.113Check: 5² = 25 and 5³ = 125, so the answer is between 2 and 3

Solving exponential equations

If both sides can be written with the same base, set the exponents equal. If not, get the power alone and take a log of both sides; the power rule brings the unknown down from the exponent.

Example 11: Same base

3^{2x − 1} = 3³27 = 3³2x − 1 = 3Same base, so the exponents are equalx = 2

Example 12: Take a log of both sides

5 · 2^x = 60Start from the original equation2^x = 12Divide both sides by 5 firstx = log 12log 2Take the log of both sides; the power rule gives x · log 2 = log 12x ≈ 3.585

Example 13: How long until money doubles at 5% continuous interest?

1000 · e^0.05t = 2000Any starting amount works; use $1000e^0.05t = 2Divide by 10000.05t = ln 2Take the natural log of both sidest ≈ 13.86About 13.86 years

Solving log equations

To solve a log equation, combine the logs into one, then rewrite it in exponential form. Always check each answer in the original equation: every log must be of a positive number. A solution that breaks this is extraneous and must be rejected.

Example 14: Rewrite in exponential form

log₃(2x + 1) = 42x + 1 = 3⁴Exponential form2x + 1 = 81x = 40Check: 2(40) + 1 = 81 > 0 ✓

Example 15: Reject an extraneous root

log₂ x + log₂(x − 2) = 3The original equation. Both logs must be of positive numbers, so x > 0 and x − 2 > 0: the domain is x > 2log₂(x(x − 2)) = 3Product rulex(x − 2) = 2³Exponential formx² − 2x − 8 = 0(x − 4)(x + 2) = 0So x = 4 or x = −2x = 4Check against the original equation's domain x > 2: 4 works, but x = −2 would need log₂(−2) and log₂(−4), which are not defined, so it is extraneous

Example 16: A quotient of logs

log(x + 3x) = log 4Quotient rule on log(x + 3) − log x = log 4x + 3x = 4Logs of equal numbers are equalx + 3 = 4xx = 1Check: x + 3 = 4 and x = 1 are both positive ✓

Graphs and asymptotes

For b > 1 the graph of y = b^x rises from left to right, passes through (0, 1) and (1, b), and gets closer and closer to the x-axis on the left without touching it: the line y = 0 is a horizontal asymptote. Its domain is all real numbers and its range is y > 0.

The graph of y = log_b x is the reflection: it passes through (1, 0) and (b, 1), and the line x = 0 is a vertical asymptote. Its domain is x > 0 and its range is all real numbers. Shifting a graph shifts its asymptote with it.

Teal: y = 2^x − 3 is y = 2^x moved down 3, so its asymptote is the dashed line y = −3. Plum: y = log₂(x + 4) is y = log₂ x moved left 4, so its asymptote is the dashed line x = −4 and its domain is x > −4.

Common mistakes

✗Writing log_b(M + N) = log_b M + log_b N.✓The product rule is about log_b(MN). There is no rule for the log of a sum.

✗Reading y = 50 · 0.92^x as a 92% decay.✓The factor is what is left: 1 − 0.92 = 0.08, an 8% decay.

✗Writing log 12log 2 as log 6.✓Change of base gives a quotient of logs, not the log of a quotient: log 12log 2 ≈ 3.585.

✗Keeping every root of a log equation.✓Check each root in the original equation; a log of zero or a negative number makes it extraneous.

✗Rounding in the middle of a calculation.✓Keep full calculator values and round only the final answer.

Practice problems

Printable worksheet (PDF)

Try each one, then open the answer to check your work.

A town of 1500 people grows 3% a year. Write a model and find the population after 10 years.
Show answer
y = 1500 · 1.03^tb = 1 + 0.03y ≈ 20161500 · 1.03¹⁰ ≈ 2015.87, about 2016 people
Find the amount when $500 is invested at 6% compounded quarterly for 3 years.
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A = 500(1 + 0.06/4)^{4 · 3}A ≈ 597.81
Write log₄ 64 = 3 in exponential form.
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4³ = 64
Evaluate log₂(1/8).
Show answer
log₂(1/8) = −32⁻³ = 1/8
Condense log₂ 40 − log₂ 5 and evaluate.
Show answer
log₂(40/5) = log₂ 8Quotient rule3
Solve 2^{x + 3} = 32.
Show answer
2^{x + 3} = 2⁵x + 3 = 5x = 2
Solve 4 · 3^x = 50. Round to 3 decimal places.
Show answer
3^x = 12.5x = log 12.5/log 3x ≈ 2.299
Solve log₅(x − 2) = 2.
Show answer
x − 2 = 5²x = 27Check: 27 − 2 = 25 > 0 ✓
Solve log₃ x + log₃(x − 8) = 2.
Show answer
x(x − 8) = 3²Product rule, then exponential formx² − 8x − 9 = 0(x − 9)(x + 1) = 0x = 9x = −1 is extraneous

History: tables that saved time

The Scottish mathematician John Napier described logarithms in a book published in 1614, Mirifici logarithmorum canonis descriptio. His aim was to turn long multiplications into additions. Napier's own logarithms were not quite the base 10 or base e logs we use today. Henry Briggs worked with him on a base 10 version, and published his first table of these common logarithms, Logarithmorum chilias prima, in 1617.

The number e came from money. According to the MacTutor History of Mathematics archive, in 1683 Jacob Bernoulli studied continuous compound interest and tried to find the limit of (1 + 1/n)ⁿ as n grows. The letter e for this number first appears in a letter Leonhard Euler wrote to Christian Goldbach in 1731. MacTutor calls the claim that Euler chose e for his own name ridiculous.

Connections

Exponential functions first appear in Functions and Their Graphs, and the log rules come straight from the rules for powers in Exponents and Radicals. Geometric sequences in Sequences and Series are exponential growth counted in whole steps.

Solving a log equation often ends in a quadratic, from Quadratic Equations, and checking for extraneous roots works just like it does in Rational Expressions and Equations.

Practice generator

Pick a level, or Mixed for a bit of everything. Answers are a single number, like 5, −2, 1/3 or 2.161. When a problem asks you to round, keep full calculator values and round only at the end. You can type x = in front, and a dollar sign or percent sign is fine.

This practice machine needs JavaScript turned on. Here are three to try on paper: (1) evaluate log base 3 of 81; (2) solve 2 to the power x equals 20, to 3 decimal places; (3) solve log base 2 of the quantity x plus 1 equals 5. Answers: (1) 4; (2) about 4.322; (3) 31.

Sources

J J O'Connor and E F Robertson, “John Napier”, MacTutor History of Mathematics, University of St Andrews: Napier biography (the 1614 book)
J J O'Connor and E F Robertson, “Henry Briggs”, MacTutor: Briggs biography (the 1617 table)
J J O'Connor and E F Robertson, “The number e”, MacTutor: The number e (Bernoulli 1683, Euler 1731)
Common Core State Standards for Mathematics: Linear, Quadratic, and Exponential Models (F-LE.A.4), Building Functions (F-BF.B.5), Seeing Structure in Expressions (A-SSE.B.3c), Interpreting Functions (F-IF.C.7e)

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