Defining the natural logarithm as an integral, proving its algebraic properties from that definition, defining e and the exponential as its inverse, giving irrational exponents a meaning, and repairing the circularity of the earlier treatment.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Integrals, Exponential Functions, and Logarithms
Objectives
Five outcomes. This section proves what Chapters 1 and 3 assumed, using nothing but Chapter 5.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 626-635 — the section these objectives are drawn from
Warm-up
Chapter 1 used exponentials and logarithms freely, and Chapter 3 differentiated them. Both took for granted that raising a number to any real power makes sense.
Discussion prompt
What does 2 to the power of the square root of two actually mean?
Hint: Repeated multiplication only defines whole-number powers.
Answer:
Repeated multiplication defines whole-number powers, roots extend that to fractions, and beyond that the usual answer is a limit of rational powers — which Chapter 1 never made precise. So the exponential was used without a definition covering the inputs it was used on.
\[ 2^{\sqrt2} = \;? \quad \text{(no rational power equals this)} \]
This section repairs that. It defines the natural logarithm as an integral, proves every property from that definition, and gets the exponential as its inverse — after which an irrational power has a precise meaning for the first time.
Concept
The natural logarithm of a positive number is defined as the area under the reciprocal from one to that number. Every property follows from that definition using Chapter 5's tools alone.
the integral definition — The natural logarithm is defined as the integral of the reciprocal from one to the input. It is a definition rather than a theorem, and everything else is proved from it.
\[ \ln x = \int_{1}^{x}\frac{dt}{t}, \quad x>0 \]
The definition is chosen precisely because Section 5.3's Part 1 then gives the derivative immediately, and the reciprocal's scale invariance gives the algebraic properties by substitution.
Figure (svg): The natural logarithm defined as the area under the reciprocal from 1
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 742-752
Section
Section 1
Concept
Defining the logarithm as the exponential's inverse requires the exponential to be defined first — and for irrational exponents it never was. Building the logarithm from an integral avoids the gap.
the circularity — Chapter 1 defined the logarithm from the exponential and assumed the exponential made sense for all real exponents, which was never established. The integral definition assumes nothing.
\[ \text{ln by integral} \;\longrightarrow\; e \;\longrightarrow\; \exp \;\longrightarrow\; a^{x} \]
The gap is not a technicality invented for this section. Nothing in Chapter 1 says what raising a number to an irrational power means, and every property used afterwards depended on it.
Figure (svg): The circularity being repaired
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 742-750 — the natural logarithm as an integral
Picture it
One assumes, one proves.
Figure (svg): The circularity being repaired
The right column starts from an integral, which Chapter 5 defined completely, and never appeals to anything unproved. That is the whole reason for the rebuild.
Worked example
Example 6.41. Two facts, straight from the definition.
\[ \text{From } \ln x=\int_{1}^{x}\frac{dt}{t}, \text{ find } \ln 1 \text{ and the derivative.} \]
Substitute x equal to 1
Why: The limits coincide.
Apply Section 5.2's property
Why: Equal limits give zero.
\[ \ln 1 = 0 \]
Apply Part 1 of the Fundamental Theorem
Why: Differentiate the area function.
State the derivative
Why: The reciprocal.
\[ \frac{1}{x} \]
Note what was used
Why: Only Chapter 5.
Figure (svg): The natural logarithm defined as the area under the reciprocal from 1
\[ \ln 1 = 0, \qquad \frac{d}{dx}\ln x = \frac1x \]
Verify: notice which result was proved and which was assumed before
Why: Section 3.9 asserted this derivative and used it throughout Chapters 3 to 5; here it is a one-line consequence of Part 1 rather than an assumption. That reversal is the section's method in miniature — the definition was chosen so that the derivative comes for free, and everything else is then built on solid ground.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 745-747
Fill the middle
The natural logarithm, built from an integral.
Fill in the blanks
\ln x = \int_1}^___\frac______
Why: The lower limit is 1, which makes the logarithm of 1 equal to zero by the equal-limits property. Any other choice would shift the function by a constant.
Worked example
Checkpoint 6.41. Two properties needed later.
\[ \text{Show } \ln \text{ is strictly increasing and takes every real value.} \]
Examine the derivative
Why: It is the reciprocal.
Conclude monotonicity
Why: Section 4.5.
Examine the behaviour at large x
Why: The area keeps growing.
Examine it near zero
Why: The area from x to 1 grows.
Apply the Intermediate Value Theorem
Why: Continuous and unbounded both ways.
Figure (svg): The solution to Worked example the logarithm is increasing and unbounded shown as a ladder of expressions, one row per legal move
\[ \ln: (0,\infty) \;\to\; (-\infty,\infty), \text{ strictly increasing} \]
Verify: see why both facts will be needed
Why: Strict monotonicity is what guarantees an inverse exists at all, and covering the whole real line is what makes that inverse defined for every real input — which is exactly what the exponential must be. So neither property is decoration: they are the hypotheses of the inverse function theorem, established before the inverse is claimed. Note the unboundedness is not obvious, since the integrand shrinks; it holds because the reciprocal shrinks too slowly.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 747-749
Trap
\[ \ln(ab)=\ln a+\ln b \text{ because that is how logarithms work} \]
Appeal to the property from Chapter 1
Why: The student uses what this section is rebuilding.
In this section the logarithm is the integral and nothing else, so its properties must be derived from that definition.
\[ \ln(ab)=\int_{1}^{ab}\frac{dt}{t}, \text{ then split and substitute} \]
Work from the definition alone
Why: Every property is a theorem here, not an assumption.
The discipline is the point of the section. Assuming a property that the definition was introduced to prove reintroduces the circularity the whole construction exists to remove.
Sorting
What the definition delivers.
Sort into buckets
Sort each statement by its status in this section.
The order matters: the derivative and monotonicity come first because the later constructions depend on them. Building in the wrong order would reintroduce exactly the circularity the section removes.
Two truths and a lie
All three are about the rebuild.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. It is the same function, arrived at without assuming anything — and the section's later work shows it has all the familiar properties. The point is the logical route, not a new object.
Prediction
Commit before reasoning.
Predict first
Why is defining the logarithm as the exponential's inverse unsatisfactory?
Correct: Because the exponential was never defined for irrational exponents.
\[ \sqrt2 \text{ is irrational: no root or power reaches it} \]
Why: Repeated multiplication handles whole numbers and roots handle fractions, but 2 to the power of the square root of 2 is neither — and the usual gesture toward a limit of rational powers requires proving that limit exists and behaves well, which Chapter 1 did not do. Since every logarithm property used afterwards rested on that, the gap was load-bearing rather than cosmetic.
Section
Section 2
Concept
The logarithm of a product splits into a sum because the reciprocal is scale invariant: substituting a scaled variable leaves the integrand's form unchanged and shifts the limits.
scale invariance — Stretching the interval of integration by a factor and rescaling the variable leaves the integral of the reciprocal unchanged, because the factor cancels between the numerator and the denominator.
\[ \int_{a}^{ab}\frac{dt}{t} = \int_{1}^{b}\frac{du}{u} \]
That cancellation is special to the reciprocal. For any other power the substitution would leave a factor behind, and no product-to-sum property would follow.
Figure (svg): The product rule for logarithms, proved by a substitution
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 746-754 — properties of the natural logarithm
Picture it
Split, substitute, recognise.
Figure (svg): The product rule for logarithms, proved by a substitution
The middle line is where the work happens. Substituting a scaled variable turns the second integral into one starting at 1, which is the definition of a logarithm again.
Worked example
Example 6.43. Split and substitute.
\[ \text{Prove } \ln(ab)=\ln a+\ln b \text{ from the definition.} \]
Write the definition
Why: The integral to ab.
\[ \int\text{ from } 1\text{ to ab of } \,dt / t \]
Split at a
Why: Section 5.2's additivity.
\[ \int 1\text{ to } a + \int a\text{ to ab} \]
Substitute in the second
Why: Let t be a times u.
\[ \,dt = a \,du \]
Simplify the integrand
Why: The factor cancels.
\[ a \,du / (a u) = \,du / u \]
Read the new limits
Why: From a to ab becomes 1 to b.
\[ \ln a + \ln b \]
Figure (svg): The product rule for logarithms, proved by a substitution
\[ \ln(ab) = \ln a + \ln b \]
Verify: identify what made the cancellation work
Why: The substitution multiplied the differential by a and the denominator by a, and those cancelled — which happens only for the reciprocal. For an integrand of t to the power negative two, say, the factors would not match and the resulting function would satisfy no such property. So the logarithm's algebra is a direct consequence of which power was integrated, which is a satisfying answer to why logarithms behave as they do.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 749-751
Fill the middle
The second integral in the product proof.
Fill in the blanks
t = au \;\Longrightarrow\; \fracu___ = \frac______ = \frac______}
Why: The factor of a appears in both the differential and the denominator, so it cancels. That cancellation is special to the reciprocal and is why only it has a product-to-sum property.
Worked example
Checkpoint 6.43. The same technique.
\[ \text{Prove } \ln(a^{r})=r\ln a \text{ for rational } r. \]
Write the definition
Why: The integral to a to the r.
\[ \int\text{ from } 1\text{ to } a ^{r}\text{ of } \,dt / t \]
Substitute
Why: Let t be u to the r.
\[ \,dt = r u ^{r - 1} \,du \]
Simplify the integrand
Why: Combine the powers.
\[ r \,du / u \]
Pull the constant out
Why: It is a number.
\[ r \times \int\text{ from } 1\text{ to } a \]
Recognise
Why: The definition again.
Figure (svg): The solution to Worked example the power rule for logarithms shown as a ladder of expressions, one row per legal move
\[ \ln(a^{r}) = r\ln a \]
Verify: note the restriction and how it is lifted
Why: The proof is stated for rational r because a to the power r must already have a meaning for the substitution to be written — and irrational powers do not yet. Once the exponential is defined in the next idea, the general power is defined so that this property holds by construction, at which point the restriction disappears. The order of the construction is doing real work here.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 751-753
Error analysis
A student proves the quotient property.
Annotate
On: \( \ln\frac{a}{b} = \ln a - \ln b \quad \text{because division undoes multiplication} \)
In this section every property is a theorem with a proof from the integral. Appealing to familiar behaviour is exactly the circularity the construction was built to avoid.
Ranking
From the definition to the property.
Put in order
Why: Step d is the crux and it depends entirely on the integrand being the reciprocal. Step b uses Section 5.2's additivity, and step e uses the definition in reverse — nothing outside Chapter 5 is needed.
Sorting
Each property has its own route.
Sort into buckets
Sort each property by what proves it.
Every route is Chapter 5 material, which is the point: the construction uses only what has been fully proved. Nothing here appeals to a property of exponentials.
Prediction
Commit before reasoning.
Predict first
Why does no other power's integral turn products into sums?
Correct: Because only for the reciprocal does the factor cancel.
\[ t=au: \; \frac{a\,du}{(au)^{n}} = a^{1-n}\frac{du}{u^{n}}, \text{ equal only if } n=1 \]
Why: Substituting a scaled variable multiplies the differential by the scale factor and the integrand by that factor raised to the power. Those cancel only when the power is negative one — for any other power a leftover factor survives, and the integral of the scaled interval is not the same as the original. So the logarithm's characteristic algebra is a consequence of which exponent was excluded from the power rule, which ties this section directly back to Section 4.10.
Section
Section 3
Concept
Since the logarithm is continuous, strictly increasing and unbounded, it takes the value one exactly once. That input is defined to be e.
the number e — The unique positive number whose natural logarithm equals one — equivalently, the point at which the area under the reciprocal from one reaches one.
\[ \ln e = 1 \quad \text{defines } e \]
The definition requires no decimal expansion, no limit of a sequence and no compound-interest story. Existence and uniqueness come from the Intermediate Value Theorem and monotonicity.
Figure (svg): Defining e as the number whose logarithm is one
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 750-758 — the number e
Picture it
A single crossing.
Figure (svg): Defining e as the number whose logarithm is one
The horizontal line at height one meets the curve exactly once, because the logarithm is strictly increasing. That single crossing is e, and nothing about its decimal expansion is needed.
Worked example
Example 6.45. Two theorems doing the work.
\[ \text{Show there is exactly one } x \text{ with } \ln x=1. \]
Note the logarithm is continuous
Why: It is differentiable.
Evaluate at 1 and at 4
Why: By estimation.
\[ \ln 1 = 0\text{ and } \ln 4 > 1 \]
Apply the Intermediate Value Theorem
Why: Section 2.4.
\[ \text{some } x\text{ gives } 1 \]
Note strict monotonicity
Why: The derivative is positive.
Conclude
Why: Exactly one such x.
Figure (svg): Defining e as the number whose logarithm is one
\[ \exists! e>0: \; \ln e = 1 \]
Verify: check the bracketing estimate
Why: The area under the reciprocal from 1 to 4 exceeds one: even the crude lower bound of three rectangles of width one and heights one half, one third and one quarter gives about 1.08. So the logarithm passes one somewhere before 4, and it is above zero at 1 — bracketing e in the interval from 1 to 4. Refining the estimate would narrow it toward 2.718, but the definition does not need that number at all.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 753-755
Fill the middle
The number e, defined by the logarithm.
Fill in the blanks
\ln e = 1
Why: The definition is that the area under the reciprocal from 1 reaches exactly one at e. No decimal expansion is involved, and existence follows from the Intermediate Value Theorem.
Worked example
Checkpoint 6.45. Consistency with Section 4.8.
\[ \text{Show this } e \text{ is the same as } \lim_{n\to\infty}(1+1/n)^{n}. \]
Take the logarithm of the expression
Why: It brings the exponent down.
\[ n \ln(1 + \frac{1}{n}) \]
Rewrite as a quotient
Why: For L'Hopital's rule.
\[ \ln(1 + \frac{1}{n}) / (\frac{1}{n}) \]
Apply the rule
Why: Section 4.8.
\[ \text{the } \lim\text{ is } 1 \]
Exponentiate
Why: Undo the logarithm.
\[ \text{the expression tends to the number whose } \log\text{ is } 1 \]
Recognise
Why: That number.
Figure (svg): The solution to Worked example recovering the familiar limit shown as a ladder of expressions, one row per legal move
\[ \lim_{n\to\infty}\left(1+\tfrac1n\right)^{n} = e \]
Verify: note which definition is primary
Why: Section 4.8 computed this limit and called the answer e, taking the exponential for granted; here e is defined by the integral and the limit is shown to reach it. The logical direction is reversed, and that reversal is the point: the limit is now a theorem about a defined object rather than the definition of an undefined one. Both give the familiar 2.71828, so nothing computational changes.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 755-757
Trap
\[ e = 2.71828\ldots \]
Define e by its digits
Why: The student treats the decimal as the definition.
A decimal expansion identifies a number but does not define it, and it gives no way to prove anything about it.
\[ \ln e = 1 \;\Longrightarrow\; e \text{ is determined exactly} \]
Define e by a property, then compute digits if wanted
Why: The property is what proofs use.
The same distinction applies to pi, defined by a ratio rather than by 3.14159. A defining property supports proofs; a decimal expansion supports only arithmetic.
Matching
Two theorems establish e.
Match the pairs
Why: Existence and uniqueness are separate claims needing separate arguments, and both trace back to the positive integrand. Nothing about e's decimal value enters at any point.
Sorting
In this section's logical order.
Sort into buckets
Sort each statement.
The fourth is the striking reclassification: Section 4.8 treated that limit as what e is, and here it is a theorem about a number already defined. Which statements are definitions and which are theorems depends on the construction chosen.
Prediction
Commit before reasoning.
Predict first
Why is defining e as 2.71828... unsatisfactory?
Correct: Because it supports no proofs.
\[ \text{a property supports proofs}; \quad \text{digits support only arithmetic} \]
Why: From the digits alone nothing can be deduced — not that the exponential is its own derivative, not that the compound-interest limit converges to it, not any of its properties. A defining property like the logarithm reaching one is what proofs actually use, and the digits are then computed from it. Pi is defined by a ratio for the same reason.
Section
Section 4
Concept
Because the logarithm is strictly increasing, continuous and onto the real line, it has an inverse defined for every real input. That inverse is the exponential function.
the natural exponential — The inverse of the natural logarithm. Its existence follows from the logarithm's strict monotonicity and its range being all of the reals.
\[ \exp = \ln^{-1}, \qquad \exp(\ln x)=x, \; \ln(\exp y)=y \]
Every property of the exponential now follows from the corresponding logarithm property, including that it is its own derivative — which Section 3.9 asserted.
Figure (svg): The exponential defined as the logarithm's inverse
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 754-760 — the exponential function
Picture it
The logarithm and its inverse.
Figure (svg): The exponential defined as the logarithm's inverse
The exponential's domain is the logarithm's range, which is the whole real line — so it is defined for every real input, including the irrational ones Chapter 1 could not handle.
Worked example
Example 6.46. Proved rather than asserted.
\[ \text{Show } \frac{d}{dx}\exp(x)=\exp(x). \]
Write the inverse relation
Why: Applying ln to exp.
\[ \ln(\exp x) = x \]
Differentiate both sides
Why: The chain rule on the left.
\[ \exp'(x) / \exp(x) = 1 \]
Solve
Why: Multiply through.
\[ \exp'(x) = \exp(x) \]
Note what was used
Why: Only the inverse relation and the chain rule.
Compare with Section 3.9
Why: It was asserted there.
Figure (svg): The exponential defined as the logarithm's inverse
\[ \exp'(x)=\exp(x) \]
Verify: trace which facts the proof rests on
Why: The proof used the logarithm's derivative, which came from Part 1 of the Fundamental Theorem, and the chain rule from Section 3.6 — both fully established. Section 3.9 stated this property and used it in dozens of places without proof; every one of those uses is now justified retrospectively. That is what the rebuild delivers: not new results, but secure ones.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 757-759
Fill the middle
The defining relation between the two functions.
Fill in the blanks
\ln(\exp y) = y
Why: Applying the logarithm to the exponential returns the input, which is what being inverses means. Differentiating this relation is how the exponential's derivative is obtained.
Worked example
Checkpoint 6.46. Every property inherited.
\[ \text{Show } \exp(a+b)=\exp(a)\exp(b). \]
Take the logarithm of the right side
Why: Apply ln.
\[ \ln(\exp a \times \exp b) \]
Apply the logarithm's product rule
Why: Proved earlier.
\[ \ln(\exp a) + \ln(\exp b) \]
Simplify by the inverse relation
Why: Each term collapses.
\[ a + b \]
Note the left side's logarithm
Why: By the inverse relation.
Conclude
Why: The logarithm is one-to-one.
Figure (svg): The solution to Worked example the exponential's algebra shown as a ladder of expressions, one row per legal move
\[ \exp(a+b)=\exp(a)\exp(b) \]
Verify: notice the pattern of every such proof
Why: The route is always the same: take logarithms, use the corresponding logarithm property, and appeal to injectivity to conclude. Every exponential identity is the mirror of a logarithm one, and each is proved by this three-step pattern. That the logarithm is one-to-one — needed in the final step — is exactly the strict monotonicity established at the start, doing work again.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 759-760
Error analysis
A student defines the exponential.
Annotate
On: \( \exp \text{ is the inverse of } \ln, \text{ obviously} \)
Two separate facts were established earlier precisely so this step would be legitimate. Skipping them leaves the exponential's domain unjustified, which is where the irrational-exponent problem came from in the first place.
Ranking
From the integral to the exponential.
Put in order
Why: Step b is what makes step c legitimate, and skipping it is the error the section exists to avoid. Every property in step e is the mirror of one already proved, obtained by the same three-line pattern.
Sorting
Each exponential identity mirrors one.
Sort into buckets
Sort each exponential property by its logarithm counterpart.
Every exponential fact is a logarithm fact in a mirror. That correspondence is what makes the construction economical: proving the logarithm's properties once gives the exponential's for free.
Prediction
Commit before reasoning.
Predict first
Why does the logarithm have an inverse defined for every real number?
Correct: Because it is one-to-one with range all the reals.
\[ \text{one-to-one} + \text{onto } \mathbb{R} \;\Longrightarrow\; \exp \text{ on all of } \mathbb{R} \]
Why: Being strictly increasing makes it one-to-one, so an inverse exists on its range; the range being the whole real line makes that inverse defined for every real input. Both facts were proved earlier for exactly this purpose, and both trace back to the reciprocal being positive and shrinking slowly. Continuity alone would not suffice, since a continuous function need not be one-to-one.
Section
Section 5
Concept
With the exponential and logarithm both established, a general power is defined as the exponential of the exponent times the logarithm of the base — which makes sense for every real exponent.
the general power — For a positive base, the power with any real exponent is defined as the exponential of that exponent times the base's logarithm. It agrees with repeated multiplication and roots on the cases those already covered.
\[ a^{x} = \exp(x\ln a), \quad a>0 \]
This is the payoff of the whole construction. The gap Chapter 1 left is filled, and every power law follows from the exponential's properties rather than being assumed.
Figure (svg): Irrational exponents finally given a meaning
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 756-762 — general logarithmic and exponential functions
Picture it
What an irrational power now means.
Figure (svg): Irrational exponents finally given a meaning
The last row matters: the new definition must agree with repeated multiplication where that already applied, or it would be defining a different operation rather than extending one.
Worked example
Example 6.47. The definition applied.
\[ \text{Give a meaning to } 2^{\sqrt2} \text{ and evaluate it.} \]
Apply the definition
Why: Exponential of exponent times logarithm.
\[ \exp(\sqrt{2} \ln 2) \]
Note both parts are defined
Why: ln 2 is an area; exp is defined on all reals.
Compute the logarithm
Why: By estimation.
\[ \text{about } 0.693 \]
Multiply by the exponent
Why: About 1.414.
\[ \text{about } 0.980 \]
Exponentiate
Why: The inverse function.
\[ \text{about } 2.665 \]
Figure (svg): Irrational exponents finally given a meaning
\[ 2^{\sqrt2} = \exp\!\left(\sqrt2\,\ln 2\right) \approx 2.665 \]
Verify: check the answer is between the neighbouring rational powers
Why: The exponent lies between 1.4 and 1.5, so the value must lie between 2 to the 1.4 and 2 to the 1.5 — about 2.639 and 2.828, and 2.665 sits between them. That check also confirms the definition extends the familiar one rather than replacing it: the new value agrees with what squeezing by rational powers would give, which is what the old hand-waving gestured at without proving.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 759-761
Fill the middle
A positive base raised to any real exponent.
Fill in the blanks
a^\ln a = \exp\!\left(x\,___\right)
Why: Both the logarithm and the exponential are defined for all the inputs involved, so the expression makes sense for every real exponent — which is exactly what the earlier treatment could not deliver.
Worked example
Checkpoint 6.47. Nothing left to assume.
\[ \text{Prove } a^{x}a^{y}=a^{x+y} \text{ from the definition.} \]
Write both factors
Why: By the definition.
\[ \exp(x \ln a) \times \exp(y \ln a) \]
Apply the exponential's product rule
Why: Proved earlier.
\[ \exp(x \ln a + y \ln a) \]
Factor
Why: Collect the logarithm.
\[ \exp((x + y) \ln a) \]
Recognise
Why: The definition again.
Note the generality
Why: x and y are any reals.
Figure (svg): The solution to Worked example the power laws follow shown as a ladder of expressions, one row per legal move
\[ a^{x}a^{y}=a^{x+y} \]
Verify: appreciate what has been achieved
Why: The law was used constantly from Chapter 1 onward and held only for the exponents its earlier definition covered — whole numbers and fractions. Now it is proved for every real exponent, from a definition that makes sense for all of them. The whole of Chapters 1 to 5's use of exponentials is retrospectively justified by this section, which is the point of putting it at the end rather than the beginning.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 761-762
Trap
\[ (-2)^{\sqrt2} = \exp\!\left(\sqrt2\,\ln(-2)\right) \]
Apply the definition to a negative base
Why: The student ignores the domain restriction.
The logarithm is defined only for positive inputs, so the expression is meaningless — and no real number is being named.
\[ a^{x} \text{ is defined for } a>0 \text{ only} \]
Check the base is positive before applying the definition
Why: The logarithm's domain is the restriction.
Negative bases genuinely do not support irrational exponents: even a negative base to the power one half is not a real number. The restriction is not an artefact of the construction but a fact about the operation.
Sorting
The base must be positive.
Sort into buckets
Sort each expression.
The last is worth noting: the cube root of negative eight is a perfectly good real number, negative two, but it comes from the root definition rather than from this one. The general power definition covers positive bases only, and other cases keep their own separate meanings.
Two truths and a lie
All three are about the general power.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The base must be positive, because the logarithm is undefined otherwise. That restriction reflects a genuine fact about the operation rather than a limitation of the method — a negative base to an irrational power is not a real number by any route.
Prediction
Commit before reasoning.
Predict first
What does this construction deliver that Chapter 1's treatment did not?
Correct: A precise meaning and proofs of what was assumed.
\[ \text{same results, secure foundations} \]
Why: No new formula appears in this section — every result was used earlier. What is new is that they are now proved, from a definition that covers every real exponent, using only Chapter 5's fully established machinery. The value is logical rather than computational, which is why the section sits at the end of the book rather than the beginning.
Comparison
Fill the blanks. One assumes, the other proves.
Comparison matrix
| Chapter 1's route | This section's route | |
|---|---|---|
| Starts from | the exponential, assumed | the logarithm, as an integral |
| Irrational exponents | never defined | defined via exp and ln |
| The derivative of ln | asserted | from the Fundamental Theorem |
| The product rule | assumed from exponent laws | proved by a substitution |
The results are identical and the logical status is not. Everything Chapters 1 to 5 did with exponentials is retrospectively justified by the right-hand column.
Pattern
Building the exponential family from nothing.
The order is forced. Step four's inverse exists only because of step three, and step five's definition needs both functions from step four — building in any other order reintroduces the circularity.
Stewart, Calculus: Early Transcendentals 8e, Appendix G — The Logarithm Defined as an Integral Appendix G, pp. A50-A56
Check
The definition.
Check your understanding
How is the natural logarithm defined in this section?
Answer: A
Why: The integral definition assumes nothing and lets every property be proved.
Check
The product rule.
Check your understanding
What proves that ln(ab) = ln a + ln b here?
Answer: A
Why: The scale factor cancels between the differential and the integrand.
Check
Irrational exponents.
Check your understanding
What does 2 to the power root 2 mean?
Answer: A
Why: The general power is defined as the exponential of the exponent times the base's logarithm.
Real world
A numerical library must evaluate a general power for arbitrary real inputs, and its authors must decide what algorithm to implement and what to guarantee about the result.
Discussion prompt
Explain what the library actually computes, why the definition matters for its correctness, and where the edge cases come from.
Hint: The library cannot multiply an irrational number of times.
Answer:
The library computes the general power exactly as this section defines it: it takes the base's natural logarithm, multiplies by the exponent, and exponentiates. There is no other route, since repeated multiplication is meaningless for a non-integer exponent, and this is why the standard library function is built on logarithm and exponential primitives.
\[ \texttt{pow}(a,x) = \exp(x\ln a) \]
The definition matters for correctness because the library's documented guarantees — that the power laws hold, that the result is continuous in both arguments, that integer exponents agree with repeated multiplication — are exactly the theorems of this section. Without them there would be nothing to test against.
The edge cases come straight from the domain restriction. A negative base has no logarithm, so the standard function returns an error there except for integer exponents, which it handles by a separate path. Zero to the zero is defined by convention rather than by the formula, since the logarithm of zero does not exist. Every awkward case in the specification traces back to the logarithm's domain.
Note also the numerical consequence: computing through logarithms loses precision when the base is near one, because the logarithm is near zero there and the multiplication amplifies its relative error. Libraries add a separate code path for that case — a practical cost of the definition, and one that only makes sense once you know what the function is actually doing.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why is the natural logarithm defined as an integral rather than as the exponential's inverse?
Correct: Because the exponential was never defined for irrational exponents.
\[ 2^{\sqrt2}: \text{ undefined before, } \exp(\sqrt2\ln 2) \text{ after} \]
Why: Repeated multiplication and roots cover whole numbers and fractions only, and Chapter 1 gestured at a limit for the rest without establishing it. Defining the logarithm from an integral assumes nothing — Chapter 5 defined integrals completely — and every property then follows as a theorem. Ease of computation is not the issue; the issue is that the earlier route rests on something unproved.
Explain it
They cannot see why anyone would define a logarithm as an area when everyone knows what a logarithm is.
Discussion prompt
In four sentences or fewer, show them the gap.
Hint: Ask them what an irrational power means.
Answer:
Ask them what 2 to the power of the square root of 2 means. Repeated multiplication needs a whole number of copies and roots need a fraction, and the square root of 2 is neither — so the usual definition of an exponential simply does not reach it.
Defining the logarithm as an area needs none of that: an integral of the reciprocal makes sense for any positive upper limit. Once you have the logarithm you get the exponential as its inverse, and only then does the irrational power get a meaning.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the motivation, ask what an irrational power means and watch the earlier definition fail. For the proofs, remember the scale factor cancels only for the reciprocal. For e, existence is the Intermediate Value Theorem and uniqueness is monotonicity. For general powers, the definition is the exponential of the exponent times the base's logarithm. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, sketch the reciprocal with the area from 1 to x shaded, and write the definition beside it with a note that this is a definition rather than a theorem. Below, write the five-step construction in order, marking beside each step which earlier result it depends on. In the middle of the page, work the product-rule proof in five lines, circling the point where the scale factor cancels and writing one sentence on why no other power would work. Beside it, sketch the logarithm crossing the height one and mark e, noting which theorem gives existence and which gives uniqueness. In the lower half, sketch the logarithm and exponential reflected in the diagonal, and write the general power definition beneath with an example of an irrational exponent evaluated. At the bottom, write in one sentence what Chapter 1 assumed that this section proves.
If your five-step list has the exponential defined before the logarithm's range is established, reorder it — that dependency is the whole reason the construction runs in this direction rather than the familiar one.
Recap
Five things, and none of them is a new formula — all are secure foundations for old ones.
| If you see | Then |
|---|---|
| An irrational exponent | Use the exponential-of-logarithm definition |
| A logarithm property to prove | Split the integral and substitute |
| A scale factor in the integrand | It cancels only for the reciprocal |
| A claim that e is 2.718 | That is a value, not a definition |
| An inverse being claimed | Check one-to-one and onto first |
| A negative base | The general power is undefined |
| A property from Chapter 1 | In this section it must be proved |
Section 6.8 uses these foundations for exponential growth and decay, and Section 6.9 closes the course with the hyperbolic functions — built from the exponential this section has just secured.
OpenStax Calculus Volume 1, §6.7 Integrals, Exponential Functions, and Logarithms §6.7, pp. 626-635 — everything on these slides traces back here
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