Contents / Precalculus / Exponential & Logarithmic Functions

Chapter 3

Exponential & Logarithmic Functions

Exponential and logarithmic rules, the natural log, logarithmic scales, and growth models.

Introduction

Exponential and logarithmic rules, the natural log, logarithmic scales, and growth models.

3.1Exponential function

Exponential functions grow (or decay) at a rate proportional to their current value. They appear everywhere in nature and finance.

Exponential Functions

Definition 3.1 (Exponential function). An exponential function has the form

f (x) = a \cdot b^{x}, b > 0, b \neq = 1, a \neq = 0.

The base $b$ (the number being repeatedly multiplied) determines the growth or decay rate; the coefficient $a$ is a vertical stretch or reflection.

The two restrictions on $b$ are not fussiness. We require $b > 0$ so that $b^{x}$ is real for every real $x$ (e.g. $(- 2)^{1/2}$ is not real), and $b \neq = 1$ because $1^{x} = 1$ is just a constant, not a genuine exponential.

Theorem 3.2 (Laws of exponents). For any base $b > 0$ and all real $x$ and $y$ ,

b^{x + y} = b^{x} \cdot b^{y}, (b^{x})^{y} = b^{x y}, b^{0} = 1.

Moreover $b^{x}$ is increasing when $b > 1$ and decreasing when $0 < b < 1$ .

The first law is the whole character of the family: adding on the exponent side is multiplying on the value side. The graph always passes through $(0, a)$ (since $b^{0} = 1$ ), has horizontal asymptote $y = 0$ , and is defined for all real $x$ .

Intuition. Exponential is "repeated multiplication by the same factor." Start with $a$ , multiply by $b$ , then by $b$ again, forever. Think of folding a piece of paper: one fold = 2 layers, two folds = 4, ten folds = 1024. The layers don't grow by a fixed amount each fold — they double, so small changes in the exponent produce huge changes in the result.

Example 3.3 (Solving a simple exponential equation). Solve $3^{2 x - 1} = 81$ .

Solution. Rewrite both sides with the same base. Since $81 = 3 \cdot 3 \cdot 3 \cdot 3 = 3^{4}$ , the equation becomes $3^{2 x - 1} = 3^{4}$ . Because $b^{x}$ is one-to-one (for $b > 0, b \neq = 1$ ), equal powers of the same base force equal exponents: $2 x - 1 = 4$ . Solve the linear equation: $2 x = 5$ , so $x = \frac{5}{2}$ . Sanity check: $3^{2 (5/2) - 1} = 3^{5 - 1} = 3^{4} = 81$ . ✓□

Example 3.4 (Reading off growth from the base). A colony starts at $200$ cells and triples every hour. Write the model and find the count after $4$ hours.

Solution. "Triples every hour" means the base is $b = 3$ and the starting value is $a = 200$ , so $P (t) = 200 \cdot 3^{t}$ with $t$ in hours. Evaluate at $t = 4$ : $P (4) = 200 \cdot 3^{4} = 200 \cdot 81 = 16200$ cells. Sanity check by stepping hour-by-hour: $200 \to 600 \to 1800 \to 5400 \to 16200$ . ✓ Each step multiplies by $3$ , as required.□

Common Mistakes

Base confusion: Writing b^x as x^b (power vs exponential).
Exponent rules with negatives: (-2)^3 = -8 but (-2)^4 = +16; fractional exponents with even roots of negative bases are not real.
Growth vs decay: For 0 < b < 1 the exponential b^x is decay (decreasing), not growth.
Forgetting b^0=1: Always the y-intercept for a* b^x is a.
Domain of exponential: Always all reals; students sometimes think like logs.
Continuous vs discrete: e^{rt} is continuous limit; (1+r)^t discrete. They are close but not identical.

Exercises for §3.1 →

3.2Logarithm

Logarithms are the inverse operation to exponentiation. They turn multiplication into addition and are essential for solving exponential equations and handling scales.

Logarithms

Definition 3.5 (Logarithm). The logarithm base $b$ of a positive number $a$ is the exponent to which $b$ must be raised to produce $a$ :

lo g_{b} a = c ⟺ b^{c} = a .

In words: a logarithm asks a question — "what power turns the base into this number?" Logarithms are defined only for a positive argument $a > 0$ and a positive base $b > 0$ with $b \neq = 1$ . (There is no real power of a positive base that produces $0$ or a negative number, which is why the argument must be positive.)

Theorem 3.6 (Laws of logarithms). For a base $b > 0$ with $b \neq = 1$ and all $x, y > 0$ ,

lo g_{b} (x y) = lo g_{b} x + lo g_{b} y, lo g_{b} \frac{x}{y} = lo g_{b} x - lo g_{b} y,

lo g_{b} (x^{y}) = y lo g_{b} x, lo g_{b} b = 1, lo g_{b} 1 = 0.

Products become sums, quotients become differences, and exponents come down in front.

Proof. Each rule is just an exponent rule read backwards. Write $x = b^{m}$ and $y = b^{n}$ , so that $m = lo g_{b} x$ and $n = lo g_{b} y$ . Because $b^{m} \cdot b^{n} = b^{m + n}$ , taking logs turns the product on the left into the sum $m + n$ on the right, which is the first rule; the other two come the same way from $b^{m} / b^{n} = b^{m - n}$ and $(b^{m})^{y} = b^{m y}$ .∎

Proposition 3.7 (Change of base). For $a > 0$ and admissible bases,

lo g_{b} a = \frac{ln a}{ln b} .

Any consistent base works in the numerator and denominator, e.g. $\frac{lo g _{10} a}{lo g _{10} b}$ .

Intuition. A log is the "how many times did I multiply" counter. If you kept multiplying by $2$ until you reached $32$ , you multiplied $5$ times: $lo g_{2} 32 = 5$ . And because the counters just add up when you multiply totals together, a log turns awkward "power towers" back into ordinary numbers you can add and subtract.

Example 3.8 (Using log properties to simplify). Simplify $lo g_{3} 9 + lo g_{3} 27 - lo g_{3} 3$ .

Solution. Direct approach — evaluate each term. Since $9 = 3^{2}$ , $lo g_{3} 9 = 2$ . Since $27 = 3^{3}$ , $lo g_{3} 27 = 3$ . And $lo g_{3} 3 = 1$ . Combine: $2 + 3 - 1 = 4$ . Check with the log rules — a sum minus a log is a product over a quotient: $lo g_{3} 9 + lo g_{3} 27 - lo g_{3} 3 = lo g_{3} (\frac{9 \cdot 27}{3}) = lo g_{3} (\frac{243}{3}) = lo g_{3} 81$ . Since $81 = 3^{4}$ , $lo g_{3} 81 = 4$ . ✓ Both methods agree.□

Example 3.9 (Solving a logarithmic equation). Solve $lo g_{2} (x - 3) = 4$ .

Solution. Rewrite in exponential form using $lo g_{b} a = c ⟺ b^{c} = a$ : here $2^{4} = x - 3$ . Since $2^{4} = 16$ , we get $x - 3 = 16$ , so $x = 19$ . Check the domain: the argument must be positive, and $x - 3 = 16 > 0$ . ✓ So $x = 19$ is valid (a solution that made the argument $\leq 0$ would have to be discarded).□

Common Mistakes

Base confusion: log_b a means "b to what power = a", not "a to what power = b". The base is always on the bottom of the log notation.
Domain of logs: Students forget that log arguments must be positive; domain of log_b is (0, ∞).
Change of base arithmetic: \log_b a = \ln a / \ln b , not \ln a - \ln b or other misapplications.
Log of sum: There is no simple log(x + y) rule; logs turn products/quotients/powers into sums, not sums into anything simpler.
Forgetting log_b b =1 and log_b 1=0: Basic values that simplify many expressions.
Negative base or arg: Logs not defined for negative or zero in reals; b must >0 ≠1.

Exercises for §3.2 →

3.3Natural logarithm

The natural log is the logarithm with base e, the special number that makes calculus clean: its derivative is itself, and ln's derivative is 1/x.

The Natural Logarithm

Definition 3.10 (Natural logarithm). The natural logarithm $ln x = lo g_{e} x$ uses the famous base $e \approx 2.718281828 \dots$ , which can be defined as the limit

e = n \to \infty lim (1 + \frac{1}{n})^{n} .

Proposition 3.11 ( $ln$ and $e^{x}$ are inverses). The natural log and the exponential base $e$ undo one another:

ln (e^{x}) = x for all real x, e^{l n x} = x for x > 0.

In calculus $e$ and $ln$ are special because $\frac{d}{d x} e^{x} = e^{x}$ (the exponential is its own derivative) and $\frac{d}{d x} ln x = \frac{1}{x}$ .

Every other logarithm is a constant multiple of the natural log: by change of base, $lo g_{b} x = \frac{ln x}{ln b}$ , so $lo g_{b} x$ is just $ln x$ scaled by the fixed factor $1/ ln b$ .

Intuition. Money in a bank that pays interest every instant grows by exactly $e^{r t}$ . The natural log is the tool that runs that continuous growth backwards — it answers "how long did this need to compound to reach that size?" $e$ is the one special growth rate at which the curve's steepness always equals its own height, so $ln$ and $e^{x}$ fit together perfectly like a lock and key.

Example 3.12 (Solving with natural logs). Solve $e^{2 x} = 7$ for $x$ .

Solution. Take the natural log of both sides: $ln (e^{2 x}) = ln 7$ . Apply $ln (e^{u}) = u$ on the left: $2 x = ln 7$ . Divide by $2$ : $x = \frac{ln 7}{2}$ . Numerically $ln 7 \approx 1.94591$ , so $x \approx 0.97295$ . Sanity check: $e^{2 (0.97295)} = e^{1.9459} \approx 7.00$ . ✓□

Example 3.13 (Isolating before taking the log). Solve $5 e^{- 0.2 t} = 2$ for $t$ .

Solution. First isolate the exponential by dividing both sides by $5$ : $e^{- 0.2 t} = \frac{2}{5} = 0.4$ . Take $ln$ of both sides: $- 0.2 t = ln 0.4$ . Since $ln 0.4 \approx - 0.91629$ , we get $- 0.2 t \approx - 0.91629$ , so $t \approx \frac{- 0.91629}{- 0.2} \approx 4.581$ . Sanity check: $5 e^{- 0.2 (4.581)} = 5 e^{- 0.9162} \approx 5 (0.40) = 2.00$ . ✓ (Note we could not take the log until the exponential was alone — $ln (5 e^{- 0.2 t}) \neq = ln 5 \cdot (- 0.2 t)$ .)□

Common Mistakes

Base confusion with e: ln means log_e , not log10 or log2; always natural in advanced math.
Domain of ln: Only positive reals; ln of negative or zero undefined in reals.
Forgetting ln(e^x) =x exactly: Simplifies many expressions.
Mixing ln with log in software: Some languages log is ln, others log10; check docs.
Change of base with ln vs log10: Either works; ln often preferred in calc because of derivative.
Approximating e or ln without calc: e≈2.718, but use exact ln when possible.

Exercises for §3.3 →

3.4Logarithmic scale

Log scales compress huge ranges into manageable visual space by making distance proportional to ratio, not difference.

Logarithmic Scales

Definition 3.14 (Logarithmic scale). A logarithmic scale spaces numbers so that equal distances correspond to equal ratios (multiplicative factors) rather than additive differences.

Classic examples:

Richter scale for earthquake magnitude (each whole number is 10× energy).
Decibel scale for sound intensity (10× power per 10 dB).
pH scale (each unit is 10× change in H+ concentration).
Astronomical magnitude and stellar brightness.

Proposition 3.15 (Exponentials are straight on a semi-log plot). If $y = a b^{x}$ then

lo g y = lo g a + x lo g b,

a straight line in the variables $x$ and $lo g y$ , with slope $lo g b$ and intercept $lo g a$ .

Proof. Take the log of both sides of $y = a b^{x}$ and use the product and power rules: $lo g y = lo g a + lo g (b^{x}) = lo g a + x lo g b$ .∎

This is why semi-log paper is useful for spotting exponential behavior in data: the eye is far better at judging straightness than at judging curvature.

Intuition. Log scales let your eyes handle numbers that differ by factors of millions without the big numbers squishing the small ones off the page. A jump from 1 to 10 looks the same size as 1,000,000 to 10,000,000.

Example 3.16 (Interpreting a decibel reading). A sound at $60$ dB is how many times more intense than one at $30$ dB? (Assume the power scale where $10$ dB $= 10 \times$ intensity.)

Solution. Because the scale is logarithmic, subtract to find the gap in dB: $60 - 30 = 30$ dB. Convert that gap to factors of $10$ : $\frac{30 dB}{10 dB per factor} = 3$ factors of $10$ . Therefore the intensity ratio is $1 0^{3} = 1000$ times. Sanity check — do not divide the readings directly: $60/30 = 2$ is wrong, because dB values are exponents, not the intensities themselves.□

Example 3.17 (Comparing earthquakes on the Richter scale). Each whole-number step on the Richter scale means $10 \times$ the amplitude. How much larger is a magnitude $7.0$ quake than a magnitude $5.5$ quake in amplitude?

Solution. Find the difference in magnitude: $7.0 - 5.5 = 1.5$ . Each unit is a factor of $10$ , so the amplitude ratio is $1 0^{1.5}$ . Compute $1 0^{1.5} = 1 0^{1} \cdot 1 0^{0.5} = 10 \cdot 10 \approx 10 \cdot 3.162 \approx 31.6$ . So the magnitude $7.0$ quake has roughly $32$ times the amplitude — a "small-looking" $1.5$ gap hides a large multiplicative jump, which is the whole point of a log scale.□

Common Mistakes

Thinking log scale is just 'log of the numbers on axes': The spacing between ticks is logarithmic.
Confusing additive vs multiplicative: +1 on Richter is *10 energy, not +1.
Forgetting that 0 may not appear or negative not shown: Many log scales start at small positive.
Misreading pH: pH7 is 10× less acidic than pH6 (lower pH more acidic).
Calculator log vs ln in scale calcs: Use consistent base; the factor is the same.
Assuming all 'log plots' are log-log: Semi-log (one axis log) straightens exponentials; log-log straightens power laws.

Exercises for §3.4 →

3.5Exponential growth

When a quantity's rate of change is proportional to its size, the result is exponential growth or decay. This leads to the differential equation dP/dt = k P whose solution is P = P0 e^{kt}.

Exponential Growth and Decay

Definition 3.18 (Exponential growth and decay). Exponential growth occurs when a quantity changes at a rate proportional to its current size,

\frac{d P}{d t} = k P with solution P (t) = P_{0} e^{k t},

where $P_{0} = P (0)$ is the starting amount and $k > 0$ . Exponential decay is the same equation with $k < 0$ (the quantity shrinks toward $0$ instead of blowing up).

Proposition 3.19 (Discrete and continuous compounding). Compounding over $t$ years at rate $r$ with $n$ compoundings per year gives

A = P (1 + \frac{r}{n})^{n t},

and taking the continuous limit $n \to \infty$ gives

A = P e^{r t} .

Proposition 3.20 (Half-life). For decay ( $k < 0$ ), the time for the quantity to halve is

t_{1/2} = \frac{ln 2}{∣ k ∣} .

It depends only on $k$ , not on the starting amount $P_{0}$ — a key feature of exponential decay.

These models appear in finance, biology, physics, and technology adoption curves.

Intuition. Exponential growth is "the bigger you are, the faster you grow." Each step multiplies by the same factor >1, so the increases themselves get larger and larger — the opposite of linear "add the same amount every time."

Example 3.21 (Continuous compounding and doubling time). A principal of $1000$ dollars is invested at $4%$ annual interest compounded continuously.

Solution. Set up the model with $P = 1000$ and $r = 0.04$ : $A (t) = 1000 e^{0.04 t}$ . To find the doubling time, set $A = 2000$ : $2000 = 1000 e^{0.04 t} ⟹ 2 = e^{0.04 t}$ . Take $ln$ of both sides: $ln 2 = 0.04 t$ , so $t = \frac{ln 2}{0.04} \approx \frac{0.693147}{0.04} \approx 17.33$ years. Sanity check with the Rule of 72: $\frac{72}{4} = 18$ years — close to $17.33$ , as expected for a quick estimate.□

Example 3.22 (Radioactive decay and half-life). A $50$ mg sample decays with rate constant $k = - 0.10$ per year. Find its half-life and the amount left after $10$ years.

Solution. Half-life uses only $∣ k ∣$ : $t_{1/2} = \frac{ln 2}{∣ k ∣} = \frac{0.693147}{0.10} \approx 6.93$ years. Amount remaining after $10$ years: $P (10) = 50 e^{(- 0.10) (10)} = 50 e^{- 1}$ . Since $e^{- 1} \approx 0.36788$ , we get $P (10) \approx 50 \cdot 0.36788 \approx 18.39$ mg. Sanity check against the half-life: after one half-life ( $\approx 6.93$ yr) about $25$ mg remain; $10$ years is a bit more than one half-life, so a value somewhat below $25$ mg — namely $18.4$ mg — is exactly what we expect. ✓□

Common Mistakes

Growth vs decay: For 0 < b < 1 the exponential b^x is decay (decreasing), not growth.
Continuous vs discrete: e^{rt} is the continuous limit; (1+r)^t is discrete annual compounding. They are close but not identical.
Half life depends on k not on P0: Time to halve is independent of starting amount.
Rule of 72 is approximation: 72/r% gives rough doubling years; exact is ln2 *100 / r .
Forgetting the e in continuous model: Many default to (1+r)^t even when problem says continuous.
Sign of k: k>0 growth, k<0 decay; |k| in half life.

Exercises for §3.5 →

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

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