Compound Interest, Continuous Compounding & Logarithms

Compound-Interest Fundamentals

  • Present value (initial principal) denoted PP.

  • Annual percentage yield (APR) denoted rr (expressed as a decimal; e.g.
    15 % ⇒ r=0.15r = 0.15).

  • Interest is reinvested (compounded) at the end of each compounding period.

  • 1-year accumulation with annual compounding:

    • A(1)=P+Pr=P(1+r)=P(1+0.15)A(1)=P+Pr = P(1+r)=P\bigl(1+0.15\bigr) when r=0.15r=0.15.

  • 2-year accumulation (still annual compounding):

    • Second-year interest is earned on BOTH the original PP and the interest from year 1.

    • A(2)=P(1+r)(1+r)=P(1+r)2A(2)=P(1+r)(1+r)=P(1+r)^2.

  • In general, after tt years with annual compounding:

    • A(t)=P(1+r)tA(t)=P(1+r)^t — exponential growth where

    • coefficient A=PA=P,

    • base B=1+rB=1+r.

Compounding mm Times per Year (Periodic Compounding)

  • Each period’s rate = rm\dfrac r m.
    – Ex: Quarterly compounding ⇒ m=4  ⇒  rate per quarter=r/4m=4\;\Rightarrow\;\text{rate per quarter}=r/4.

  • Number of periods in tt years = mtmt.

  • General periodic-compounding formula:

    • A(t)=P(1+rm)mt\displaystyle A(t)=P\left(1+\frac{r}{m}\right)^{mt}.

  • Special case annual compounding: m=1m=1 gives A(t)=P(1+r)tA(t)=P(1+r)^t again.

Worked Example 1 – $2000 at 12.6 % Compounded Monthly

  • Given: P=2000 ,  r=12.6%=0.126 ,  m=12 (monthly) ,  t=2.5 yrP=2000\,,\;r=12.6\%=0.126\,,\;m=12\text{ (monthly)}\,,\;t=2.5\text{ yr}.

  • Plug into periodic formula:
    A(2.5)=2000(1+0.12612)12×2.5=2000 (1.0105)30≈2793.!97.\displaystyle A(2.5)=2000\Bigl(1+\tfrac{0.126}{12}\Bigr)^{12\times2.5}=2000\,(1.0105)^{30}\approx 2793.!97.

  • Calculator entry tip:
    2000*(1+0.126/12)^(12*2.5).

Building an Explicit Exponential Model & Goal-Time Estimate

  • Model form: A(t)=abtA(t)=ab^{t}.
    – Here a=P=2000a=P=2000.
    – b=(1+0.126/12)12=1.010512≈1.126b=(1+0.126/12)^{12}=1.0105^{12}\approx1.126.

  • Final model: A(t)=2000 (1.0105)12t\displaystyle A(t)=2000\,(1.0105)^{12t}.

  • Estimate when A(t)=5000A(t)=5000 (table or calculator).
    – Table shows t=8t=8 is first year where A≥5000A\ge5000.
    – So investment reaches $5000 during the 8th year.

Radioactive-Decay Example (Carbon-14)

  • Carbon-14 amount modeled by
    C(t)=A (0.999879)t,t=years.C(t)=A\,\bigl(0.999879\bigr)^{t},\quad t=\text{years}.

  • A fossil contains 0.5 0.5\,g after t=50,000t=50{,}000 yr.
    Set C(50,000)=0.5C(50{,}000)=0.5 and solve for the original amount AA:
    A=0.50.99987950,000≈2.12 g.\displaystyle A=\frac{0.5}{0.999879^{50{,}000}}\approx2.12\,\text{g}.

  • Illustrates exponential decay that happens continuously in nature, not in discrete steps.

Discrete vs Continuous Compounding

  • Banking/finance: compounding happens on discrete schedule (daily, monthly, etc.).

  • Natural processes (radioactive decay, population growth of bacteria, etc.) act continuously.

  • Conceptual leap: let compounding frequency m→∞m\to\infty to model “continuous” compounding.

Continuous Compounding & Euler’s Number ee

  • Toy experiment: invest $1 for 1 yr at 100 % interest, compounded mm times/year:
    A(1)=(1+1m)m.A(1)=\left(1+\frac1m\right)^{m}.

  • As m→∞m\to\infty, the limit of (1+1/m)m\bigl(1+1/m\bigr)^{m} approaches e≈2.718281828…e\approx2.718281828\ldots.

  • ee (Euler’s number) is the limiting factor for continuous growth; crucial in calculus and finance.

Continuous-Compounding Formula & Syntax

  • For principal PP, annual rate rr (decimal), time tt (years):
    A(t)=P ert.\boxed{\displaystyle A(t)=P\,e^{rt}}.

  • Growth if r>0r>0; decay if r<0r<0.

  • Calculator shortcuts:
    – TI-83/84: e^(...) or EXP(...).
    – Excel: =EXP(rt).

Worked Example 2 – Growth (Bank Account)

  • P=10,000 ,  r=0.06 ,  continuous.P=10{,}000\,,\;r=0.06\,,\;\text{continuous}.

  • Balance function: A(t)=10,000 e0.06t.A(t)=10{,}000\,e^{0.06t}.

  • After t=5t=5 yr: A(5)=10,000 e0.3≈13,498.59.A(5)=10{,}000\,e^{0.3}\approx13{,}498.59.

Worked Example 3 – Continuous Decline (Stock)

  • Same P=10,000P=10{,}000 but declining at 6 % continuously ⇒ r=−0.06r=-0.06.

  • A(t)=10,000 e−0.06t.A(t)=10{,}000\,e^{-0.06t}.

  • After 5 yr: A(5)=10,000 e−0.3≈7,408.18.A(5)=10{,}000\,e^{-0.3}\approx7{,}408.18.

Introduction to Logarithms

  • Definition: y=log⁡b(x)  ⟺  by=x.\displaystyle y=\log_{b}(x)\iff b^{y}=x.
    – “The power you raise bb to in order to obtain xx.”

  • Examples:

    • log⁡2(8)=3    ∵  23=8.\log_{2}(8)=3\;\;\because\;2^{3}=8.

    • log⁡3(1/27)=−3    ∵  3−3=1/27.\log_{3}\bigl( 1/27 \bigr)=-3\;\;\because\;3^{-3}=1/27.

  • Domain restrictions: b>0,  b≠1,  x>0b>0,\;b\neq1,\;x>0.
    – log⁡<em>3(−9)\log<em>{3}(-9) undefined (cannot reach a negative with positive base). – log⁡</em>1(2)\log</em>{1}(2) impossible (base 1 always returns 1).

Special Bases

  • Common logarithm: log⁡10(x)\log_{10}(x), usually written just log⁡(x).\log(x).

  • Natural logarithm (Napierian): log⁡e(x)\log_{e}(x), written ln⁡(x).\ln(x).

  • Quick evaluations:

    • log⁡(10,000)=4\log(10{,}000)=4 because 104=10,00010^{4}=10{,}000.

    • ln⁡(e)=1  ;  ln⁡(1)=0.\ln(e)=1\;;\;\ln(1)=0.

Change-of-Base Formula & Calculator Tips

  • log⁡b(a)=log⁡(a)log⁡(b)=ln⁡(a)ln⁡(b).\displaystyle \log_{b}(a)=\frac{\log(a)}{\log(b)}=\frac{\ln(a)}{\ln(b)}.

  • Allows any base to be computed on a standard calculator.
    – Example: log⁡11(9)=ln⁡(9)ln⁡(11)≈0.91631.\log_{11}(9)=\dfrac{\ln(9)}{\ln(11)}\approx0.91631.

Solving Exponential Equations with Logs

  • Equation 5−x=1255^{-x}=125 ⇒ rewrite as log:
    −x=log⁡<em>5(125)⇒x=−log⁡</em>5(125)=−ln⁡125ln⁡5=−3.-x=\log<em>{5}(125) \Rightarrow x=-\log</em>{5}(125)=-\frac{\ln125}{\ln5}=-3.

  • Equation 3  2x−1=63^{\;2x-1}=6:
    2x−1=log⁡<em>3(6)  ⇒  x=log⁡</em>3(6)+12≈1.3155.2x-1=\log<em>{3}(6)\;\Rightarrow\;x=\dfrac{\log</em>{3}(6)+1}{2}\approx1.3155.

Logarithmic Function (General Form)

  • Simplest descriptive form:
    f(x)=a ln⁡(x)+c,a≠0.\displaystyle f(x)=a\,\ln(x)+c, \quad a\neq0.

  • More general statement: any f(x)=log⁡b(x)+cf(x)=\log_{b}(x)+c for positive base b≠1b\neq1 is “logarithmic.”

  • These will be graphed and analyzed (shifts, stretches, asymptotes) in next lecture.

Ethical, Historical & Practical Notes

  • John Napier (17th century) introduced logarithms to simplify enormous hand calculations; enabled Kepler’s celestial computations.

  • ee surfaces naturally whenever growth/decay is continuous, linking finance, physics, biology, chemistry and information theory.

Looking Ahead

  • Upcoming topics:
    – Graphing log functions & identifying asymptotes.
    – Logarithmic identities (product, quotient, power rules).
    – Applications: investment doubling time, radioactive half-life, logarithmic regression.