Compound Interest, Continuous Compounding & Logarithms

Compound-Interest Fundamentals

• Present value (initial principal) denoted $P$.

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

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

• 1-year accumulation with annual compounding:

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

• 2-year accumulation (still annual compounding):

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

  • $A(2)=P(1+r)(1+r)=P(1+r)^2$.

• In general, after $t$ years with annual compounding:

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

  • coefficient $A=P$,

  • base $B=1+r$.

Compounding $m$ Times per Year (Periodic Compounding)

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

• Number of periods in $t$ years = $mt$.

• General periodic-compounding formula:

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

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

Worked Example 1 – $2000 at 12.6 % Compounded Monthly

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

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

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

Building an Explicit Exponential Model & Goal-Time Estimate

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

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

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

Radioactive-Decay Example (Carbon-14)

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

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

• 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 $m\to\infty$ to model “continuous” compounding.

Continuous Compounding & Euler’s Number $e$

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

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

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

Continuous-Compounding Formula & Syntax

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

• Growth if r>0; decay if r<0.

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

Worked Example 2 – Growth (Bank Account)

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

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

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

Worked Example 3 – Continuous Decline (Stock)

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

• $A(t)=10{,}000\,e^{-0.06t}.$

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

Introduction to Logarithms

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

• Examples:

  • $\log_{2}(8)=3\;\;\because\;2^{3}=8.$

  • $\log_{3}\bigl( 1/27 \bigr)=-3\;\;\because\;3^{-3}=1/27.$

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

Special Bases

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

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

• Quick evaluations:

  • $\log(10{,}000)=4$ because $10^{4}=10{,}000$.

  • $\ln(e)=1\;;\;\ln(1)=0.$

Change-of-Base Formula & Calculator Tips

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

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

Solving Exponential Equations with Logs

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

• Equation $3^{\;2x-1}=6$:
  $2x-1=\log<em>{3}(6)\;\Rightarrow\;x=\dfrac{\log</em>{3}(6)+1}{2}\approx1.3155.$

Logarithmic Function (General Form)

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

• More general statement: any $f(x)=\log_{b}(x)+c$ for positive base $b\neq1$ 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.

• $e$ 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.