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AdvancedGrades 10–1125 min45 / 82

Logarithms, their properties and the logarithmic function

The definition of the logarithm and the basic identity, common (log) and natural (ln) logarithms, the product, quotient and power rules, change of base, the function y = logₐx, comparing logarithms, and applications: pH, decibels, the earthquake scale and doubling time.

Check yourself
In this lesson you will learn
  • Evaluate logarithms from the definition and use the basic logarithmic identity
  • Simplify expressions with the product, quotient and power rules and the change-of-base formula
  • Know the graph and properties of y = logₐx and compare logarithms
  • Use logarithmic scales (pH, decibels, magnitude) and the doubling-time formula

In the previous lesson we got stuck on 2ˣ = 5: 2² = 4 is too small, 2³ = 8 too big, so x lies between 2 and 3, yet it cannot be written as a “nice” number. Aysel asks a similar question: at 10% a year, how many years does it take her deposit to double? In 1.1ⁿ = 2 the unknown is again in the exponent. For such numbers mathematicians created a new name and symbol, the logarithm: x = log₂5 ≈ 2.32 and n = log₁.₁2 ≈ 7.27.

In this lesson we study the logarithm from its definition: how to compute it, what its properties are, what the graph of y = logₐx looks like and where logarithms are used in real life. Logarithmic equations and inequalities are solved in the next lesson, “Logarithmic equations and inequalities”.

The definition and the basic identity

Definition
Logarithm

The logarithm of a positive number b to base a (logₐb) is the number c such that aᶜ = b; here a > 0 and a ≠ 1. In other words, a logarithm answers the question “to what power must a be raised to get b?” For example, log₂8 = 3, because 2³ = 8.

logₐb = c ⇔ aᶜ = b (a > 0, a ≠ 1, b > 0)
where:
  • athe base: a > 0, a ≠ 1
  • bthe argument: b > 0
  • cthe value of the logarithm, any real number (it may be negative)

A logarithm is an exponent. The same equality can be read in two ways: 2³ = 8 ⇔ log₂8 = 3. Special cases: logₐ1 = 0, because a⁰ = 1; logₐa = 1, because a¹ = a.

The reasons come from the exponential function. Since aᶜ is always positive, negative numbers and zero have no logarithm: log₂(−8) and log₂0 make no sense. For a = 1 we have 1ᶜ = 1, so for log₁5, say, no c exists. The key consequence: the expression under a logarithm must always be positive, and that is why we will find domains in the next lesson.

Evaluating from the definition

Evaluate:
1) log₂32
2) log₃(1/9)
3) log₅√5
4) log₁₀0.001
5) log₀.₅8
6) log₄8

Show solution
1) 2⁵ = 32 ⇒ log₂32 = 5.
2) 3⁻² = 1/9 ⇒ log₃(1/9) = −2.
3) √5 = 5^(1/2) ⇒ log₅√5 = 1/2.
4) 10⁻³ = 0.001 ⇒ log₁₀0.001 = −3.
5) 0.5⁻³ = 2³ = 8 ⇒ log₀.₅8 = −3.
6) 4 = 2², 8 = 2³: (2²)ᶜ = 2³ ⇒ 2c = 3 ⇒ log₄8 = 3/2.
Definition
Common and natural logarithms

The logarithm to base 10 is the common logarithm, written log b (some books write lg b): log b = log₁₀b. The logarithm to base e ≈ 2.718 is the natural logarithm: ln b = logₑb. The log and ln keys of a calculator compute exactly these two: log 1000 = 3, log 0.01 = −2, ln e = 1, ln 1 = 0.

a^(logₐb) = b logₐ(aᶜ) = c
where:
  • athe base: a > 0, a ≠ 1
  • ba positive number
  • cany real number

The basic logarithmic identity: raising a to the logarithm of b gives back b. Raising to a power and taking a logarithm “undo” each other, like addition and subtraction.

Using the basic identity

Evaluate:
1) 2^(log₂7)
2) 3^(2 + log₃5)
3) 10^(1 + log 3)
4) 25^(log₅3)
5) ln e³ and e^(ln 5)

Show solution
1) Straight from the identity: 7.
2) Split the power into two factors: 3² · 3^(log₃5) = 9 · 5 = 45.
3) 10¹ · 10^(log 3) = 10 · 3 = 30.
4) 25 = 5²: 25^(log₅3) = (5^(log₅3))² = 3² = 9.
5) ln e³ = 3, e^(ln 5) = 5.

Laws of logarithms

A logarithm is an exponent, so its laws follow from the laws of powers. Let x = aᵐ and y = aⁿ, that is, m = logₐx and n = logₐy. Then xy = aᵐ⁺ⁿ, x/y = aᵐ⁻ⁿ and xᵖ = a^(pm). Reading these equalities through the definition gives the three main laws: multiplication turns into adding logarithms, division into subtracting them, and raising to a power into multiplying.

logₐ(xy) = logₐx + logₐy
where:
  • x, ypositive numbers

The logarithm of a product is the sum of the logarithms of the factors. It also reads from right to left: a sum of logarithms with the same base is the logarithm of the product.

The logarithm of a product

1) log₆4 + log₆9
2) log 25 + log 4
3) log₃18 + log₃1.5
4) Write log₂(8x) in terms of log₂x (x > 0).

Show solution
1) log₆(4 · 9) = log₆36 = 2.
2) log(25 · 4) = log 100 = 2.
3) log₃(18 · 1.5) = log₃27 = 3.
4) log₂8 + log₂x = 3 + log₂x.
logₐ(x / y) = logₐx − logₐylogₐ(x / y) = logₐx − logₐy
where:
  • x, ypositive numbers

The logarithm of a quotient is the logarithm of the dividend minus the logarithm of the divisor. Special case: logₐ(1/y) = −logₐy.

The logarithm of a quotient

1) log₃54 − log₃2
2) log 5000 − log 5
3) log₂24 − log₂3
4) log₅(1/125)

Show solution
1) log₃(54 ÷ 2) = log₃27 = 3.
2) log(5000 ÷ 5) = log 1000 = 3.
3) log₂(24 ÷ 3) = log₂8 = 3.
4) log₅(1/125) = −log₅125 = −3.
logₐ(xᵖ) = p · logₐx log_(aᵏ) x = (1/k) · logₐxlogₐ(xᵖ) = p · logₐx log_(aᵏ) x = (1/k) · logₐx
where:
  • xa positive number
  • pany real number
  • kthe exponent of the base, k ≠ 0

The exponent of the argument comes out in front as a factor, the exponent of the base comes out as its reciprocal: log₈x = log_(2³) x = ⅓ · log₂x.

The logarithm of a power

1) log₂(1/32)
2) log₃√27
3) 2 log₅10 − log₅4
4) Write log₉(x²) in terms of log₃x (x > 0).

Show solution
1) 1/32 = 2⁻⁵ ⇒ −5 · log₂2 = −5.
2) √27 = 3^(3/2) ⇒ (3/2) · log₃3 = 3/2.
3) 2 log₅10 = log₅100, then the quotient: log₅(100 ÷ 4) = log₅25 = 2.
4) 9 = 3²: log₉(x²) = ½ · log₃(x²) = ½ · 2 log₃x = log₃x.

Change of base

Let c = logₐb, that is, aᶜ = b. Take the logarithm to base k of both sides: c · logₖa = logₖb, so c = logₖb / logₖa. This formula lets us compute any logarithm with the log or ln key of a calculator and bring logarithms with different bases to one base.

logₐx = logₖx / logₖa logₐx = 1 / logₓa logₐk · logₖx = logₐxlogₐx = logₖx / logₖa logₐx = 1 / logₓa logₐk · logₖx = logₐx
where:
  • kthe new base: k > 0, k ≠ 1 (often 10 or e)
  • xa positive number; in the second formula also x ≠ 1

The change-of-base formula and two of its consequences. On a calculator: logₐx = log x / log a = ln x / ln a.

Changing the base

1) log₄8
2) log₂5 (with a calculator)
3) log₂3 · log₃4
4) log₈2
5) If log₂3 = a, express log₆12 in terms of a.

Show solution
1) Go to base 2: log₂8 / log₂4 = 3/2.
2) log 5 / log 2 ≈ 0.69897 / 0.30103 ≈ 2.32. Check: 2² = 4 < 5 < 8 = 2³.
3) The “chain”: log₂3 · log₃4 = log₂4 = 2.
4) 1 / log₂8 = 1/3.
5) log₆12 = log₂12 / log₂6 = (log₂4 + log₂3) / (log₂2 + log₂3) = (2 + a) / (1 + a).
  1. 1
    Use one base

    If the bases differ, bring all of them to one base (often 2, 3, 5 or 10) with the change-of-base formula.

  2. 2
    Move coefficients inside

    Move a number in front of a logarithm inside as an exponent: 2 log₅10 = log₅100.

  3. 3
    Combine

    Collect a sum into the logarithm of a product and a difference into the logarithm of a quotient.

  4. 4
    Evaluate and check

    Evaluate the result from the definition or with the basic identity, then check the answer with an estimate.

Exam-style expressions

Evaluate:
1) (log 8 + log 18) / (2 log 2 + log 3)
2) log₃5 · log₂₅27
3) 3^(log₉16)

Show solution
1) Numerator: log(8 · 18) = log 144. Denominator: log 4 + log 3 = log 12. Since 144 = 12², log 144 / log 12 = 2 log 12 / log 12 = 2.
2) log₂₅27 = log_(5²)(3³) = (3/2) · log₅3. Since log₃5 · log₅3 = 1, the product is 3/2.
3) log₉16 = log_(3²)(4²) = log₃4 ⇒ 3^(log₃4) = 4.
LawExample
logₐ1 = 0, logₐa = 1log₇1 = 0, log₇7 = 1
a^(logₐb) = b2^(log₂7) = 7
logₐ(xy) = logₐx + logₐylog₆4 + log₆9 = log₆36 = 2
logₐ(x/y) = logₐx − logₐylog₃54 − log₃2 = log₃27 = 3
logₐ(xᵖ) = p · logₐxlog₂(1/32) = log₂2⁻⁵ = −5
logₐx = logₖx / logₖalog₉27 = log₃27 / log₃9 = 3/2
logₐx = 1 / logₓalog₈2 = 1 / log₂8 = 1/3
All the main laws in one table. They hold only inside the domain: arguments positive, bases positive and not equal to 1.
Check yourself: fill in the gap
  1. 1.log₂32 =
  2. 2.log 0.01 =
  3. 3.log₃(1/27) =
  4. 4.log₄2 =
  5. 5.5^(log₅7) =
  6. 6.log₆2 + log₆3 =

The logarithmic function y = logₐx

The function y = aˣ takes each positive value only once, so it has an inverse: we solve for x, x = logₐy, and swap the letters: y = logₐx. Graphs of inverse functions are symmetric about the line y = x, so the properties of the logarithmic function mirror those of the exponential one, with x and y swapped.

y = logₐx (a > 0, a ≠ 1, x > 0)
where:
  • athe base
  • xthe argument, positive numbers only

The inverse of y = aˣ. Domain — (0, +∞), range — all real numbers.

  • The graph passes through (1, 0) and (a, 1), because logₐ1 = 0 and logₐa = 1.
  • For a > 1 the function is increasing, for 0 < a < 1 decreasing.
  • The y-axis (x = 0) is a vertical asymptote: as x approaches zero, the graph goes down (a > 1) or up (0 < a < 1) without bound.
  • Sign: if a and x are on the same side of 1, logₐx > 0; if they are on different sides, logₐx < 0.
Interactive
Loading simulation…
The graphs of y = aˣ and y = logₐx (written here as ln x / ln a) are mirror images in the line y = x. Make a smaller than 1 and both functions become decreasing. For a = 1 the logarithm is meaningless.
a > 1: logₐx₁ < logₐx₂ ⇔ 0 < x₁ < x₂; 0 < a < 1: logₐx₁ < logₐx₂ ⇔ x₁ > x₂ > 0
where:
  • x₁, x₂the positive arguments being compared

Comparing logarithms. If the bases differ, compare each logarithm with a “benchmark” number such as 0 or 1: 1 = logₐa, 2 = logₐa².

Comparing logarithms and finding a domain

1) log₂5 and log₂7
2) log₀.₃5 and log₀.₃7
3) log₂3 and log₃2
4) log₃10 and 2
5) Find the sign of log₀.₅3.
6) Find the domain of y = log(4 − x²).

Show solution
1) The base 2 > 1, the function is increasing, 5 < 7 ⇒ log₂5 < log₂7.
2) The base 0.3 < 1, the function is decreasing, 5 < 7 ⇒ log₀.₃5 > log₀.₃7.
3) The bases differ, so compare with 1: log₂3 > log₂2 = 1, log₃2 < log₃3 = 1 ⇒ log₂3 > log₃2.
4) 2 = log₃9 and 10 > 9 ⇒ log₃10 > 2.
5) The base 0.5 < 1 and the argument 3 > 1 are on different sides of 1 ⇒ log₀.₅3 < 0.
6) 4 − x² > 0 ⇒ x² < 4 ⇒ −2 < x < 2.

Logarithms in real life

A logarithmic scale describes a quantity that varies over a huge range with convenient numbers: a difference of one unit on the scale means a 10-fold change in the quantity (on the decibel scale this is 10 dB). Such scales are used in chemistry, acoustics and seismology.

pH = −log[H⁺] L = 10 · log(I / I₀)pH = −log[H⁺] L = 10 · log(I / I₀)
where:
  • [H⁺]the hydrogen ion concentration (mol/L)
  • Lthe sound level in decibels (dB)
  • I, I₀the sound intensity and the threshold of hearing

In pure water [H⁺] = 10⁻⁷ mol/L, so pH = 7; pH < 7 means acidic, pH > 7 basic. Earthquake magnitude is a common logarithm too: one unit more means 10 times the amplitude recorded by a seismograph and about 32 times the energy released.

pH, decibels and magnitude

1) A solution has [H⁺] = 10⁻³ mol/L. What is its pH? Is it acidic or basic?
2) Find the pH when [H⁺] = 2 · 10⁻⁵ mol/L (log 2 ≈ 0.301).
3) How many times larger is [H⁺] in a solution with pH = 4 than in one with pH = 6?
4) A sound is 10⁶ times more intense than the threshold of hearing. What is its level? What is the level if the intensity grows another 100 times?
5) How many times larger is the amplitude of a magnitude 7 earthquake than that of a magnitude 5 one?

Show solution
1) pH = −log 10⁻³ = 3 < 7, acidic.
2) pH = −log(2 · 10⁻⁵) = −(log 2 − 5) = 5 − 0.301 ≈ 4.7.
3) [H⁺] = 10⁻⁴ and 10⁻⁶: 10⁻⁴ ÷ 10⁻⁶ = 100 times.
4) L = 10 · log 10⁶ = 60 dB; with another 100 times the intensity, 10 · log 10⁸ = 80 dB (+20 dB).
5) The difference is 2 units ⇒ 10² = 100 times.

Back to Aysel’s question: 1.1ⁿ = 2. Taking the common logarithm of both sides and using the power law gives n · log 1.1 = log 2: the unknown “comes down” from the exponent and an ordinary linear equation is left. Taking logarithms of both sides will be one of the main tools for solving equations in the next lesson.

(1 + p/100)ⁿ = 2 ⇒ n = log 2 / log(1 + p/100) ≈ 72 / p(1 + p/100)ⁿ = 2 ⇒ n = log 2 / log(1 + p/100) ≈ 72 / p
where:
  • pthe growth rate per period, in %
  • nthe number of periods needed to double

The doubling time. The approximate rule of 72 (n ≈ 72/p) works well for small rates and lets you estimate in your head.

Doubling time

1) At 10% a year, how many years does it take Aysel’s deposit to double?
2) How many years at 8% a year? Check with the rule of 72.
3) A number of bacteria grows by 50% every hour. After how many hours has it grown 10 times?

Show solution
1) n = log 2 / log 1.1 ≈ 0.30103 / 0.04139 ≈ 7.27: by the end of year 8 the balance has more than doubled. Rule of 72: 72 ÷ 10 = 7.2.
2) n = log 2 / log 1.08 ≈ 9.0 years; 72 ÷ 8 = 9, a match.
3) 1.5ᵗ = 10 ⇒ t = log 10 / log 1.5 = 1 / log 1.5 ≈ 1 / 0.176 ≈ 5.7 hours.

Key points

  • logₐb = c ⇔ aᶜ = b (a > 0, a ≠ 1, b > 0); a^(logₐb) = b, logₐ1 = 0, logₐa = 1.
  • log b = log₁₀b is the common logarithm, ln b = logₑb the natural logarithm.
  • logₐ(xy) = logₐx + logₐy, logₐ(x/y) = logₐx − logₐy, logₐ(xᵖ) = p · logₐx; there is no rule for the logarithm of a sum.
  • Change of base: logₐx = logₖx / logₖa = log x / log a; logₐx = 1 / logₓa.
  • y = logₐx is the inverse of y = aˣ: domain (0, +∞), the graph passes through (1, 0), increasing for a > 1 and decreasing for 0 < a < 1.
  • Logarithmic scales: pH = −log[H⁺], L = 10 · log(I/I₀); doubling time n = log 2 / log(1 + p/100) ≈ 72/p.

Check yourself

12 questions. Every correct answer earns XP.

1 / 12
Which of these expressions is defined?