Day 20 — Exponential growth & decay

Day 20 — Exponential growth & decay

Stage II · concept day
Goal: Model \(a\cdot r^{t}\) growth/decay; compute doubling time and half-life; compare linear vs exponential tables; use compound interest; treat logs as inverses of exponentials to solve for time.

Why this matters

Moore’s-law style growth, epidemic toy models, cache miss penalties that multiply, and compound interest are exponential. Algorithms that are \(2^n\) become infeasible quickly—the table comparison makes this visceral. Logs convert multiply-growth into additive time equations.

Theory

Exponential functions

For base \(b>0\), \(b\neq 1\): \[f(x)=b^{x}.\]

• Domain \(\mathbb{R}\) (for real teaching; constructive details aside).
  
• Range \((0,\infty)\).
  
• If \(b>1\): increasing (growth). If \(0<b<1\): decreasing (decay).
  
• \(b^{0}=1\); \(b^{1}=b\); \(b^{-x}=1/b^{x}\).

General model: \[A(t)=A_0\, r^{t}\] or \(A(t)=A_0 e^{kt}\) with \(k=\ln r\) when using continuous base \(e\).

Often \(r=1+i\) for growth rate \(i\), or \(r=1-\delta\) for decay.

Doubling time

If \(A(t)=A_0\cdot 2^{t/T_d}\), then \(T_d\) is the doubling period.
If \(A(t)=A_0 b^{t}\) with \(b>1\), solve \(b^{T}=2\): \(T=\log_b 2=\dfrac{\ln 2}{\ln b}\).

Half-life

If \(A(t)=A_0\cdot (1/2)^{t/T_h}\), then \(T_h\) is half-life.
For \(A(t)=A_0 e^{-kt}\) (\(k>0\)), \(T_h=\dfrac{\ln 2}{k}\).

Linear vs exponential tables

\(t\)	Linear \(2t+1\)	Exp \(2^{t}\)
0	1	1
1	3	2
2	5	4
3	7	8
4	9	16
5	11	32

Linear: constant additive change. Exponential: constant multiplicative factor.

Rule of thumb: any exponential with base \(>1\) eventually dominates any polynomial (Stage VIII makes precise).

Compound interest

Principal \(P\), annual rate \(r\) (decimal), compounded \(n\) times per year for \(t\) years: \[A=P\biggl(1+\frac{r}{n}\biggr)^{nt}.\]

Continuous compounding: \(A=Pe^{rt}\).

Effective annual rate: \(\bigl(1+\frac{r}{n}\bigr)^{n}-1\).

Logs invert exponentials

\[y=b^{x}\quad\Leftrightarrow\quad x=\log_b y.\]

Solve \(A_0 r^{t}=A\) by \(t=\log_r(A/A_0)=\dfrac{\ln(A/A_0)}{\ln r}\).

Exponential equations (review+)

Same base: \(b^{f(x)}=b^{g(x)}\Rightarrow f(x)=g(x)\).
Different bases: take log, or rewrite with prime bases.

Exponent laws (active recall)

For \(b>0\), \(b\neq 1\) and real exponents as taught: \[ b^{x+y}=b^x b^y,\quad b^{xy}=(b^x)^y,\quad (ab)^x=a^x b^x,\quad b^{-x}=\frac{1}{b^x}. \] Logs invert: \(\log_b(xy)=\log_b x+\log_b y\), \(\log_b(x^r)=r\log_b x\), change of base \(\log_b x=\frac{\ln x}{\ln b}\).

Why exponential beats polynomial (sketch)

Fix \(r>1\) and degree \(d\). Ratios \(\frac{r^{n+1}/(n+1)^d}{r^n/n^d}=r\bigl(\frac{n}{n+1}\bigr)^d\to r>1\), so eventually the sequence \(r^n/n^d\) increases and \(\to\infty\). Thus \(r^n\) grows faster than \(n^d\). (Stage VIII: Big-O formalizes this.)

Compound frequency comparison

For fixed \(P,r,t>0\), the function \(n\mapsto P\bigl(1+\frac{r}{n}\bigr)^{nt}\) increases with \(n\) toward \(Pe^{rt}\). More frequent compounding → slightly more money, with continuous as the upper envelope.

Worked examples

Example 1 — Evaluate

\(3^{4}=81\), \(2^{-3}=\frac{1}{8}\), \((\frac{1}{2})^{3}=\frac{1}{8}\).

Example 2 — Growth model

Bacteria \(A_0=500\), double every \(3\) hours: \(A(t)=500\cdot 2^{t/3}\). At \(t=9\): \(500\cdot 2^{3}=4000\).

Example 3 — Doubling time

\(A=100\cdot (1.05)^{t}\). Doubling: \(1.05^{T}=2\), \(T=\frac{\ln 2}{\ln 1.05}\approx\frac{0.693}{0.0488}\approx 14.2\) years.

Example 4 — Half-life

Isotope half-life \(8\) days, \(A_0=40\) g: \(A(t)=40\cdot(\frac{1}{2})^{t/8}\). After \(24\) days: \(40\cdot\frac{1}{8}=5\) g.

Example 5 — Compound interest

\(P=1000\), \(r=0.06\), \(n=12\), \(t=2\): \(A=1000(1.005)^{24}\approx 1127.49\).

Example 6 — Continuous

\(1000 e^{0.06\cdot 2}=1000 e^{0.12}\approx 1127.5\) (similar).

Example 7 — Solve for time

\(500\cdot 1.03^{t}=800\). \(1.03^{t}=1.6\). \(t=\frac{\ln 1.6}{\ln 1.03}\approx 15.9\).

Example 8 — Linear vs exp

After \(10\) steps: linear \(2\cdot 10+1=21\) vs \(2^{10}=1024\).

Example 9 — Decay factor

Lose \(20\%\) per year: multiply by \(0.8\) each year. After \(5\) years: \(A_0(0.8)^{5}\).

Example 10 — Log inverse

\(f(x)=2^{x}\), \(f^{-1}(x)=\log_2 x\). Compose: \(2^{\log_2 9}=9\).

Example 11 — Effective rate

\(r=12\%\) compounded monthly: \((1.01)^{12}-1\approx 0.1268=12.68\%\) effective.

Example 12 — Equation

\(4^{x}=8^{x-1}\Rightarrow (2^2)^{x}=(2^3)^{x-1}\Rightarrow 2x=3x-3\Rightarrow x=3\).

Example 13 — Table second look

Exponential \(3^{t}\): ratios \(\frac{A(t+1)}{A(t)}=3\) constant.

Example 14 — Bits growth

Address space \(2^{n}\) grows by factor \(2\) when \(n\) increases by \(1\)—exponential in bit width.

Example 15 — Substitution exponential equation

\(2^{2x}-2^{x}-6=0\). Set \(u=2^{x}>0\): \(u^2-u-6=0\), \((u-3)(u+2)=0\), \(u=3\) (reject \(-2\)). \(2^{x}=3\), \(x=\log_2 3\).

Example 16 — Rule of 72

At \(r=0.06\), rule of 72 gives \(72/6=12\) years; exact \(T=\ln 2/\ln 1.06\approx 11.9\). Good approximation.

Example 17 — Half-life to continuous \(k\)

\(T_h=8\) days: \(e^{-k\cdot 8}=\frac12\), \(k=\frac{\ln 2}{8}\). Then \(A(t)=A_0 e^{-(\ln 2)t/8}\).

Example 18 — Comparing early table values

\(100n\) vs \(2^n\): at \(n=1..9\), linear can win; at \(n=10\), \(1000\) vs \(1024\)—already close; \(n=20\): \(2000\) vs \(>10^6\). Never trust small-\(n\) tables alone for asymptotics.

Example 19 — Password entropy

Charset size \(62\), length \(8\): \(62^8\) possibilities; bits \(\log_2(62^8)=8\log_2 62\approx 8\cdot 5.95=47.6\) bits. Each extra character adds \(\log_2 62\) bits.

Example 20 — Same growth two writings

\(A=100\cdot 4^{t}=100\cdot (2^2)^{t}=100\cdot 2^{2t}=100\cdot e^{t\ln 4}\). Discrete base \(4\) per unit time equals continuous \(k=\ln 4\).

Exercises

Easy

1. Evaluate \(5^{3}\), \(10^{-2}\), \((\frac{2}{3})^{2}\).
2. \(A(t)=20\cdot 2^{t}\): find \(A(0)\), \(A(3)\).
3. Write a model: starts at \(100\), triples every period \(t\) in periods.
4. Half-life \(10\) h, start \(80\) mg: amount after \(30\) h.
5. Solve \(2^{x}=32\).

Medium

6. Doubling time for \(6\%\) annual growth (annual compound once): \(1.06^{T}=2\).
7. \(P=2000\), \(r=4\%\), compounded quarterly, \(t=5\) years: write expression for \(A\) (compute if desired).
8. Compare after \(t=8\): \(10+3t\) vs \(10\cdot (1.2)^{t}\) qualitatively which larger.
9. Solve \(300 e^{0.02 t}=900\) for \(t\).
10. If half-life is \(T_h\), show \(A(2T_h)=A_0/4\).
11. Effective annual rate for \(r=0.05\) compounded monthly.
12. Rewrite \(A=50\cdot 4^{t}\) as \(A=50\cdot 2^{?}\).
13. Decay \(15\%\) per step for \(6\) steps: overall factor.

Hard

14. Prove that \(f(x)=b^{x}\) for \(b>1\) is injective; conclude logs well-defined.
15. Derive doubling time formula \(T=\ln 2/\ln b\) for \(A_0 b^{t}\).
16. Show continuous model \(Pe^{rt}\) is limit idea of \((1+\frac{r}{n})^{n}\to e^{r}\) (state without full limit proof).
17. Solve \(2^{2x}-2^{x}-6=0\) with \(u=2^{x}\).
18. Compare polynomial \(n^{5}\) and exponential \(1.1^{n}\) for \(n=10,50,100\) with a calculator after estimating by hand which wins eventually.
19. If \(A(t)=A_0 e^{kt}\) and \(A(3)=2A_0\), find \(k\) and half-life/doubling.
20. Prove \((b^{x})(b^{y})=b^{x+y}\) using exponent laws (Day 8 link).

Challenge / CS-flavored

21. Password entropy: space size \(N\), “bits” \(\log_2 N\). If \(N\) multiplies by \(64\), bits increase by \(?\)
22. Algorithm A: \(1000n\) ops; B: \(2^{n}\). For which \(n\) is B already larger? (try small \(n\) table)
23. Cache: miss penalty multiplies latency by \(100\) each level of memory—exponential cost stack idea.
24. Version growth: semantic version major bumps as rare exponential compatibility breaks—metaphor only.
25. Solve for bits \(n\) in \(2^{n}\ge 10^{12}\).

Continuous growth rate

If \(A=A_0 e^{kt}\), the instantaneous relative growth rate is \(k\) (calculus interpretation later).
For discrete \(A_{n+1}=(1+i)A_n\), relative growth per step is \(i\).

Log scales

Plotting \(y\) on a log scale linearizes pure exponential data: \(\log A(t)=\log A_0 + t\log r\). Slope estimates \(\log r\).

Comparing models

Given data table, look at first differences (linear?) vs ratios \(A_{t+1}/A_t\) (exponential?). Mixed models exist (affine \(A+b\) then exp)—out of scope.

Rule of \(72\) (finance heuristic)

Doubling time years \(\approx 72/(\text{percent rate})\) for modest rates—approximation to \(\ln 2/\ln(1+r)\approx 0.69/r\).

Extra exercises

26. Derive \(T_{1/2}=\ln 2/k\) from \(A_0 e^{-kt}\).
    
27. \(P=5000\) at \(3\%\) continuous for \(10\) years: write \(A\).
    
28. Solve \(4\cdot 3^{x}=36\).
    
29. If compute capacity doubles every \(18\) months, factor over \(6\) years.
    
30. Show \(b^{x}=e^{x\ln b}\) for \(b>0\) (definition bridge).

CS connection

Exponential time algorithms, cryptographic key space \(2^{n}\), compound metrics, and exponential backoff (\(2^{k}\) wait) are daily CS. Logs turn growth into linear equations for “how long until.” Always distinguish \(2^{n}\) from \(n^{2}\).

Common pitfalls

Pitfall	What to do instead
\((1+r)^{t}\) vs \(1+rt\)	Compound vs simple interest
Half-life as subtract half once	Multiplicative \((\frac12)^{t/T}\) each period
Logs of growth without dividing by \(A_0\)	\(t=\log(A/A_0)/\log r\)
Base \(b=1\) as exponential interest	Degenerate constant
Confusing percent and growth factor	\(+5\%\) means \(\times 1.05\)
Claiming linear eventually beats exp	Opposite for base \(>1\)

Fully worked extra examples

E15 — Same-base rewrite.
\((\frac{1}{8})^{x}=2^{x+3}\). Left \((2^{-3})^{x}=2^{-3x}\). So \(-3x=x+3\), \(-4x=3\), \(x=-\frac34\).

E16 — Continuous vs annual.
\(P=1000\), \(r=0.05\), \(t=10\): annual \(1000(1.05)^{10}\); continuous \(1000 e^{0.5}\). Continuous is slightly larger—limit of compounding frequency.

E17 — Half-life chain.
From \(A=A_0(\frac12)^{t/T}\), take log: \(\ln(A/A_0)=-\frac{t}{T}\ln 2\), so \(t=T\frac{\ln(A_0/A)}{\ln 2}\).

E18 — Linear never catches pure exp eventually.
\(1000n\) vs \(2^{n}\): at \(n=20\), \(20000\) vs about \(1\) million. At \(n=10\), \(10000\) vs \(1024\)—linear still larger; crossover later. Always check large \(n\) for exp base \(>1\).

E19 — Compound factor table.
\((1.01)^{12}\approx 1.1268\) (monthly \(12\%\) APR). \((1.12)^{1}=1.12\) simple annual. Difference is compound effect within the year.

End-of-day synthesis problems

S1. Model: start \(200\), grow \(8\%\) per period. Write \(A(t)\); find \(t\) with \(A(t)=500\) in log form.

S2. Half-life \(6\) h, \(A_0=64\): amount at \(t=0,6,12,18,24\).

S3. Compound: \(P=1500\), \(r=0.05\), \(n=4\), \(t=3\) — expression for \(A\).

S4. Table compare \(5+2t\) vs \(5\cdot 2^{t}\) for \(t=0..6\); when does exp overtake?

S5. Solve \(9^{x}=27^{x/2+1}\).

S6. Doubling time for continuous rate \(k=0.07\): \(T=\ln 2/k\).

S7. Effective annual rate for \(12\%\) compounded monthly.

S8. Bits: if key space multiplies by \(2^{10}\), how many bits of security added?

Deep dive — Discrete vs continuous growth

Discrete	Continuous
\(A_{t+1}=r A_t\)	\(A'(t)=k A(t)\) (calculus)
\(A(t)=A_0 r^{t}\)	\(A(t)=A_0 e^{kt}\)
\(r=e^{k}\) bridge	\(k=\ln r\)
Compound \(n\) times/year	\(n\to\infty\) limit \(e^{rt}\)

Solving for time always uses a log: the log is the inverse “how many multiplications.” In CS, exponential backoff, geometric sequences of array capacity (2^k growth), and \(O(2^n)\) algorithms are the same multiplicative structure.

Synthesis

Exponential models multiply by a constant each period. Doubling time and half-life are special level-crossing times solved with logs. Compound interest is discrete exponential; continuous compounding is the \(e^{rt}\) limit. Linear vs exponential tables teach humility about long horizons. Logs undo exponentials—always isolate the exponential first.

S9. Solve \(2^{2x}-5\cdot 2^{x}+4=0\) via \(u=2^{x}\).

S10. Population \(A_0=1000\) grows \(2\%\) per year: years until \(2500\) (log form + approximate).

S11. Show that if capacity doubles every \(18\) months, then in \(6\) years it multiplies by \(2^{4}=16\).

S12. Prove injectivity of \(b^{x}\) for \(b>1\) using: if \(b^{x}=b^{y}\) then \(b^{x-y}=1\) and \(x-y=0\).

Checkpoint

• Write \(A_0 r^{t}\) models for growth/decay
  
• Compute doubling time / half-life setups
  
• Contrast linear vs exponential tables
  
• Apply compound interest formula
  
• Solve for time using logarithms
  
• S1 and S4 qualitative mastery

Write two takeaways in your own words.

Quick reference card

Model	Formula
Discrete growth	\(A_0 r^{t}\)
Doubling period	\(T=\log_r 2\)
Half-life model	\(A_0(1/2)^{t/T_h}\)
Compound interest	\(P(1+r/n)^{nt}\)
Continuous	\(Pe^{rt}\)
Solve for time	\(t=\log_r(A/A_0)\)
Log inverse	\(y=b^{x}\Leftrightarrow x=\log_b y\)

Exp eventually beats linear when base \(>1\); tables can mislead early.

Tomorrow

Day 21 — Summation notation. \(\sum\) notation, index shifts, linearity, closed forms for \(\sum i\), \(\sum i^2\), geometric sums, telescoping, double sums, nested-loop connection.