Continuity and Differentiability · Lesson 4 of 8
Exponential and Logarithmic Functions
“Explore two inverse families of functions, understand their graphs and laws, and differentiate natural exponentials and logarithms.”
• Distinguish a fixed power xⁿ from an exponential bˣ and compare their growth. • Describe exponential and logarithmic domains, ranges, intercepts, and graph behaviour. • Explain logarithms as inverse operations and distinguish common from natural logarithms. • Derive and apply change of base and the product, quotient, and power laws. • Differentiate eˣ and log x and use these derivatives in compositions. • Check positivity and other domain restrictions before logarithmic calculations.
A changing exponent creates a new kind of growth
In xⁿ the exponent n is fixed and the base x changes. In bˣ the base b is fixed and the exponent x changes. These are different families of functions. Comparing their graphs introduces a new kind of growth, and studying the inverse operation leads to logarithms.
For x>1, increasing a positive integer exponent makes xⁿ larger. For example, when x moves from 1 to 2, x¹⁰ moves from 1 to 1024, while x¹⁵ moves from 1 to 32768. This comparison depends on x>1: between zero and one, higher positive powers have smaller values. We should not turn a positive-axis growth observation into a claim about every real input.
An increasing exponential such as 10ˣ eventually outgrows every fixed positive integer power xⁿ. The word eventually matters: we are comparing sufficiently large inputs, not asserting an inequality at every positive x. To see the scale, at x=1000 the value of x¹⁰⁰ is (10³)¹⁰⁰=10³⁰⁰, whereas 10ˣ is 10¹⁰⁰⁰. The exponential is already vastly larger in that comparison. A proof of the full eventual-growth statement is beyond the present treatment.
For a fixed real base b>1, the function f(x)=bˣ is an increasing exponential function. Its input x occurs in the exponent, and its domain is all real numbers.
The output bˣ is always positive, so the range is (0,∞). At zero the output is 1 because b⁰=1; hence every such graph passes through (0,1). As x increases the output increases. For increasingly negative x, bˣ becomes positive and arbitrarily close to zero, but it never equals zero at a finite input. The x-axis is approached without being reached.
| Feature | bˣ, with b>1 |
|---|---|
| Domain | All real numbers |
| Range | Positive real numbers |
| Fixed point on every graph | (0,1) |
| Direction | Increasing from left to right |
| Behaviour for large negative x | Approaches zero while staying positive |
The exponential with base 10 is called the common exponential. Another important base is the number e, approximately 2.71828. One way it appears is through the series 1+1/1!+1/2!+1/3!+⋯, whose value lies between 2 and 3. The symbol ! denotes factorial: for example 3!=3·2·1. The function eˣ is called the natural exponential, and it has an especially simple derivative.
The notation exp(u) means e raised to u. For example, exp(x²) is the natural exponential with exponent x², and exp(log x) means e raised to log x. It is useful for keeping a long exponent readable in a sentence; it introduces the same function, not a new operation.
Make a small value table for 2ˣ, 3ˣ, and 4ˣ at x=−2, −1, 0, 1, and 2, then sketch the three curves. Notice their common point (0,1), their order for positive inputs, and how that order reverses for negative inputs. None of the curves crosses the x-axis.
Logarithms answer an exponent question
If 2³=8, we can ask which exponent on 2 produces 8. The answer is 3, written log₂8=3. A logarithm records the exponent needed to produce a positive number from a fixed base. It is therefore an inverse operation to exponentiation.
For b>1 and a>0, log_b a is the real number x for which bˣ=a. In symbols, log_b a=x if and only if bˣ=a.
Problem
Evaluate log₂8, log₁₀10000, log₅625, and log₂₅625.
- 1.Since 2³=8, log₂8=3. The logarithm is the exponent 3, not the product 2·8.
- 2.Since 10⁴=10000, log₁₀10000=4.
- 3.The same positive number 625 equals 5⁴ and 25². Consequently log₅625=4 while log₂₅625=2.
- 4.The base is part of the definition. Changing the base changes the exponent needed to reach a fixed positive output.
For a fixed b>1, log_b x is a function from positive real inputs to all real outputs. Its domain is (0,∞), because b raised to a real power is always positive. Its range is the whole real line, because any real exponent can occur. It passes through (1,0), since b⁰=1, and increases as x increases.
As x approaches zero through positive values, log_b x decreases without a lower bound. The y-axis is approached but cannot be included in the graph. Logarithms of zero and negative numbers are not real-valued logarithms in this chapter. That restriction must be checked before using any logarithm law.
Base-10 logarithms are called common logarithms. Base-e logarithms are called natural logarithms and are written ln x. Throughout these lessons, log x without a written base means the natural logarithm ln x, matching the convention used in the chapter. It does not mean base 10 here.
The inverse relationship exchanges input and output. If eᵘ=v, then log v=u. Thus a point (u,v) on y=eˣ becomes (v,u) on y=log x. Exchanging the coordinates reflects a graph in the line y=x, explaining the paired exponential and logarithm curves.
Problem
Is exp(log x)=x true for every real x? Compare it with log(eˣ)=x.
- 1.The expression log x is real only for x>0. Therefore exp(log x) is not a real expression at zero or at a negative x.
- 2.For x>0, put u=log x. By the definition of natural logarithm, eᵘ=x. Hence exp(log x)=x on precisely that positive domain.
- 3.By contrast eˣ is positive for every real x, so log(eˣ) is defined for every real x and equals x.
- 4.An inverse identity always carries domain information. The two compositions have different domains even though both simplify to the same symbol x.
Why logarithm laws work
Logarithm laws come from the rules for exponents. They turn multiplication of positive quantities into addition of their exponents. Understanding that origin makes the laws easier to remember and shows why we cannot apply them to sums or to nonpositive factors.
Let α=log_b p and β=log_b q for positive p and q. Then p=bᵅ and q=bᵝ, so pq=bᵅbᵝ=bᵅ⁺ᵝ. The exponent needed to obtain pq is α+β, which proves the product law. Dividing the powers similarly gives the quotient law. Raising p=bᵅ to a power r gives pʳ=bʳᵅ, producing the power law for positive p.
For change of base, let α=log_a p, β=log_b p, and γ=log_b a, with a,b>1 and p>0. The definitions give aᵅ=p, bᵝ=p, and bᵞ=a. Substitution yields p=(bᵞ)ᵅ=bᵞᵅ, so β=γα. Since γ≠0, dividing gives the change-of-base formula.
Problem
Simplify log₂(8·16/4), and decide whether log₂(8+8)=log₂8+log₂8.
- 1.The product and quotient laws give log₂8+log₂16−log₂4=3+4−2=5.
- 2.A direct check gives 8·16/4=32=2⁵, confirming the result.
- 3.For the sum, log₂(8+8)=log₂16=4, while log₂8+log₂8=3+3=6.
- 4.Thus the proposed addition law is false. The product law applies to products, not to sums that happen to contain positive terms.
For example, (−2)(−3)=6 has a real logarithm, but log(−2)+log(−3) is not a real expression. The product law as used here requires each factor to be positive. Always check each logarithm you introduce, not only the final product.
Differentiating natural exponentials and logarithms
The number e has a special role in calculus: differentiating eˣ leaves the function unchanged. Its inverse, the natural logarithm, has derivative 1/x on its positive domain. We use these derivative theorems here without proving the limiting results behind them, then combine them with the chain rule.
If the exponent is u(x), the chain rule gives d(eᵘ)/dx=eᵘu′. If the logarithm’s argument is a positive function u(x), it gives d(log u)/dx=u′/u. The derivative 1/u belongs to the outer logarithm, and u′ measures how its input changes. Positivity of u is a condition on the original function as well as on the calculation.
Problem
Differentiate e⁻ˣ and sin(log x), x>0.
- 1.For e⁻ˣ, the exponent u=−x has derivative −1. Hence d(e⁻ˣ)/dx=e⁻ˣ(−1)=−e⁻ˣ.
- 2.For sin(log x), the outer sine gives cos(log x) and the inner logarithm gives 1/x.
- 3.Their product is cos(log x)/x. The argument log x explains why this function is restricted to x>0.
Problem
Differentiate cos⁻¹(eˣ) and exp(cos x).
- 1.For cos⁻¹(eˣ), the outer inverse cosine contributes −1/√(1−(eˣ)²), and the inner exponential contributes eˣ.
- 2.Thus y′=−eˣ/√(1−e²ˣ). The function is real for x≤0, while the displayed finite derivative applies for x<0.
- 3.For exp(cos x), differentiate the exponent cos x to get −sin x. Therefore y′=−sin x·exp(cos x).
- 4.The second function is defined and differentiable for all real x. Similar-looking combinations of the same functions can have very different domains.
Problem
Differentiate exp(sin⁻¹x) and log(cos(eˣ)).
- 1.For exp(sin⁻¹x), the exponential factor is unchanged and the inverse sine contributes 1/√(1−x²). The result is exp(sin⁻¹x)/√(1−x²) on −1<x<1.
- 2.The function itself is also defined at x=±1, but that does not give a finite derivative there.
- 3.For log(cos(eˣ)), the outer logarithm requires cos(eˣ)>0. Its derivative is [−sin(eˣ)·eˣ]/cos(eˣ).
- 4.Thus y′=−eˣtan(eˣ) on the intervals where the original logarithm is defined. A derivative formula on some other interval would not define the original function there.
Problem
Differentiate y=√(exp(√x)) for x>0.
- 1.Because the exponential is positive, √(exp(√x))=exp(√x/2). This simplification reduces one layer.
- 2.The derivative of the exponent √x/2 is 1/(4√x).
- 3.The chain rule therefore gives y′=exp(√x/2)/(4√x). Direct differentiation of all three layers gives the same result.
- 4.The function has a value at x=0, but the stated derivative calculation is for x>0, where the root’s derivative is finite.
Problem
Differentiate cos x/log x and log(log x).
- 1.For the quotient, the function requires x>0 and log x≠0, so x≠1. Use u′v−uv′ over v².
- 2.The derivative is [−sin x·log x−cos x/x]/(log x)². The numerator has two terms because both numerator and denominator change.
- 3.For log(log x), the outer argument log x must be positive, so x>1.
- 4.Its derivative is [1/(log x)]·(1/x)=1/[x log x], valid on that domain. Merely checking x>0 would overlook the outer logarithm.
Check your understanding
Use the graphs, inverse relationship, and derivative rules together. Several distractors ignore a domain condition or forget an inner derivative. The practice also asks you to justify logarithm laws from exponents rather than recite them.
Quiz
For b>1, which describes the range of bˣ?
What does log₅625=4 mean?
For which real x is exp(log x)=x a defined real identity?
Which logarithm law is valid for p,q>0?
What is the derivative of exp(x³)?
What is the real domain of log(log x)?
Where cos(eˣ)>0, what is the derivative of log(cos(eˣ))?
For x>0, what is d[√(exp(√x))]/dx?
Practice Problems
- Prepare value tables and sketches of 2ˣ, 3ˣ, and 4ˣ. Describe their shared point, positive-input order, negative-input order, and behaviour toward the x-axis.
- Compare 10ˣ with x¹⁰⁰ at x=1000. Explain why eventual dominance is different from an inequality at every input.
- Use exponent definitions to derive the product and change-of-base laws. Then derive the quotient and power laws for positive arguments.
- Give the domains and simplify where valid: exp(log x), log(eˣ), log(x²)−2log x, and log(log x).
- Differentiate eˣ/sin x, exp(sin⁻¹x), and exp(x³). State the domain restrictions relevant to each derivative.
- Differentiate sin(tan⁻¹(e⁻ˣ)) by following all three layers. Simplify the answer if useful.
- Differentiate log(cos(eˣ)), stating the positivity condition on its argument.
- Differentiate eˣ+exp(x²)+exp(x³)+exp(x⁴)+exp(x⁵). Explain why the five exponential terms have different multipliers.
- Differentiate √(exp(√x)) for x>0, both by simplifying first and by applying the chain rule to the original layers.
- Differentiate log(log x) and cos x/log x. Explain why the second expression excludes x=1 while the first requires x>1.
- Differentiate cos(log x+eˣ) for x>0. Show the derivative of the complete inner sum.
- The points (0,1) and (1,e) lie on y=eˣ. Find their reflected points on y=log x and explain why reflection in y=x produces them.
Key Takeaways
• In a power xⁿ the exponent is fixed; in an exponential bˣ the exponent varies. • For b>1, bˣ maps all real inputs to positive outputs, while log_b x reverses that relationship. • Natural logarithm means base e; log without a base in these lessons denotes ln. • Logarithm laws follow from exponent laws and require valid positive arguments. • The basic derivatives are (eˣ)′=eˣ and (log x)′=1/x, and compositions require inner-derivative factors. • Function domains and derivative domains must both be checked in nested logarithmic, radical, and inverse trigonometric expressions.