The Central Idea
In an exponential process, the same multiplicative change repeats over equal intervals.
If a growing quantity doubles every \(D\) units of time, then
\[ \boxed{A(t)=A_0\,2^{t/D}}. \]If a decaying quantity has half-life \(H\), then
\[ \boxed{A(t)=A_0\left(\frac12\right)^{t/H}}. \]In both formulas, the exponent counts how many doubling intervals or half-life intervals have elapsed.
What Is Doubling Time?
The doubling time of an exponential growth process is the amount of time required for the quantity to become twice its current value.
If a population doubles every \(6\) years, then a value of \(P\) becomes \(2P\) after \(6\) years.
After another \(6\) years, it becomes
\[ 4P. \]Repeated Doubling
Suppose an initial quantity is \(A_0\) and doubles once every interval.
\[ A_0,\quad2A_0,\quad4A_0,\quad8A_0,\quad16A_0,\ldots \]These values can be written
\[ A_0\,2^0,\quad A_0\,2^1,\quad A_0\,2^2,\quad A_0\,2^3,\quad A_0\,2^4,\ldots \]The exponent counts the number of doublings.
Doubling-Time Form
If one doubling requires \(D\) units of time, then after time \(t\) the number of doubling intervals is
\[ \frac{t}{D}. \]Therefore,
\[ \boxed{A(t)=A_0\,2^{t/D}}. \]Why the Exponent Is \(t/D\)
Suppose a quantity doubles every \(4\) years.
After \(4\) years,
\[ \frac{t}{D}=\frac44=1, \]so the quantity is multiplied by
\[ 2^1=2. \]After \(8\) years,
\[ \frac84=2, \]so the multiplier is
\[ 2^2=4. \]Example: Population Doubling
Suppose a population begins at \(8000\) and doubles every \(6\) years.
\[ \boxed{P(t)=8000\,2^{t/6}}. \]After \(18\) years,
\[ P(18)=8000\,2^{18/6} =8000\,2^3 =8000(8). \]Therefore,
\[ \boxed{P(18)=64000}. \]Fractional Numbers of Doublings
The quotient \(t/D\) does not need to be an integer.
If a quantity doubles every \(10\) years, then after \(5\) years the model uses
\[ 2^{5/10}=2^{1/2}=\sqrt2. \]The exponential model therefore describes values between complete doublings as well as values at the doubling times themselves.
What Is Half-Life?
The half-life of an exponential decay process is the time required for the quantity to fall to half its current value.
If a substance has a half-life of \(5\) years, then an amount \(A\) becomes \(A/2\) after \(5\) years.
Repeated Halving
Suppose the initial amount is \(A_0\).
\[ A_0,\quad \frac{A_0}{2},\quad \frac{A_0}{4},\quad \frac{A_0}{8},\quad \frac{A_0}{16},\ldots \]Equivalently,
\[ A_0\left(\frac12\right)^0,\quad A_0\left(\frac12\right)^1,\quad A_0\left(\frac12\right)^2,\quad A_0\left(\frac12\right)^3,\ldots \]Half-Life Form
If one half-life lasts \(H\) units of time, then the number of half-lives elapsed after time \(t\) is
\[ \frac{t}{H}. \]Therefore,
\[ \boxed{A(t)=A_0\left(\frac12\right)^{t/H}}. \]Why Half-Life Is Constant
In an ideal exponential decay model, the same fraction remains during equal time intervals.
\[ A_0\longrightarrow\frac{A_0}{2} \]during the first half-life, and
\[ \frac{A_0}{2}\longrightarrow\frac{A_0}{4} \]during the next.
The absolute amount lost becomes smaller, but the multiplicative factor remains \(\frac12\).
Example: Repeated Half-Lives
Suppose a substance begins with \(640\) grams and has a half-life of \(3\) years.
\[ \boxed{A(t)=640\left(\frac12\right)^{t/3}}. \]After \(15\) years,
\[ \frac{15}{3}=5 \]half-lives have passed.
\[ A(15)=640\left(\frac12\right)^5=20. \]Therefore,
\[ \boxed{A(15)=20\text{ grams}}. \]Half-Life Does Not Mean Half Disappears Only Once
If a quantity has a half-life of \(4\) hours, the process continues after the first \(4\) hours:
\[ 100\%\longrightarrow50\%\longrightarrow25\%\longrightarrow12.5\%\longrightarrow\cdots \]The remaining amount is halved again during each additional half-life.
Doubling and Halving as Multiplicative Time Scales
Doubling time and half-life describe the same exponential idea from opposite directions.
Doubling uses factor \(2\), while half-life uses factor \(\frac12\).
In each case, the time scale tells us how long it takes to apply one specified multiplicative factor.
From a General Growth Model to Doubling Time
Suppose
\[ A(t)=A_0b^t,\qquad b>1. \]At doubling time \(D\),
\[ A(D)=2A_0. \]Therefore,
\[ 2A_0=A_0b^D. \]Divide by \(A_0\):
\[ \boxed{2=b^D}. \]Solving for Doubling Time
From
\[ 2=b^D, \]take logarithms:
\[ \ln2=D\ln b. \]Therefore,
\[ \boxed{D=\frac{\ln2}{\ln b}}. \]This converts a one-unit growth factor into a doubling time.
Doubling Time from a Percentage Growth Rate
If the growth rate per period is \(r\), then
\[ b=1+r. \]Therefore,
\[ \boxed{D=\frac{\ln2}{\ln(1+r)}}. \]Example: Doubling Time at \(5\%\) Growth
A \(5\%\) annual growth rate gives factor
\[ 1.05. \]Therefore,
\[ D=\frac{\ln2}{\ln(1.05)}. \]Numerically,
\[ \boxed{D\approx14.21\text{ years}}. \]Example: Doubling Time at \(8\%\) Growth
For \(8\%\) growth per period,
\[ b=1.08. \]Thus
\[ D=\frac{\ln2}{\ln(1.08)}. \]Therefore,
\[ \boxed{D\approx9.01\text{ periods}}. \]Faster Growth Means Shorter Doubling Time
A larger growth factor causes a quantity to reach twice its current value sooner.
\[ \boxed{\text{larger growth rate}\longrightarrow\text{shorter doubling time}}. \]The relationship is not linear, but the direction is important.
From a General Decay Model to Half-Life
Suppose
\[ A(t)=A_0b^t,\qquad0Therefore,
\[ \frac{A_0}{2}=A_0b^H. \]Divide by \(A_0\):
\[ \boxed{\frac12=b^H}. \]Solving for Half-Life
Starting with
\[ \frac12=b^H, \]take logarithms:
\[ \ln\left(\frac12\right)=H\ln b. \]Therefore,
\[ \boxed{H=\frac{\ln(1/2)}{\ln b}}. \]Both logarithms are negative, so their quotient is positive.
Half-Life from a Percentage Decay Rate
If a quantity loses fraction \(r\) during each period, its remaining factor is
\[ b=1-r. \]Therefore,
\[ \boxed{H=\frac{\ln(1/2)}{\ln(1-r)}}. \]Example: Half-Life with \(12\%\) Decay
A \(12\%\) annual decay rate gives remaining factor
\[ 0.88. \]Therefore,
\[ H=\frac{\ln(1/2)}{\ln(0.88)}. \]Thus
\[ \boxed{H\approx5.42\text{ years}}. \]Example: Half-Life with \(20\%\) Decay
A \(20\%\) decay rate gives remaining factor
\[ b=0.80. \]Therefore,
\[ H=\frac{\ln(1/2)}{\ln(0.80)}. \]Hence,
\[ \boxed{H\approx3.11\text{ periods}}. \]Faster Decay Means Shorter Half-Life
A quantity that loses a larger fraction during each period reaches half its starting value sooner.
\[ \boxed{\text{larger decay rate}\longrightarrow\text{shorter half-life}}. \]Continuous Exponential Growth
A continuous exponential growth model can be written
\[ A(t)=A_0e^{kt},\qquad k>0. \]The parameter \(k\) controls the continuous rate of exponential growth.
Doubling Time for Continuous Growth
At doubling time \(D\),
\[ 2A_0=A_0e^{kD}. \]Divide by \(A_0\):
\[ 2=e^{kD}. \]Take natural logarithms:
\[ \ln2=kD. \]Therefore,
\[ \boxed{D=\frac{\ln2}{k}}. \]Example: Continuous Doubling Time
Suppose
\[ A(t)=A_0e^{0.04t}. \]Then
\[ D=\frac{\ln2}{0.04}. \]Thus,
\[ \boxed{D\approx17.33\text{ time units}}. \]Continuous Exponential Decay
A continuous decay model is often written
\[ A(t)=A_0e^{-kt},\qquad k>0. \]The negative exponent causes the exponential factor to decrease as time increases.
Half-Life for Continuous Decay
At the half-life \(H\),
\[ \frac{A_0}{2}=A_0e^{-kH}. \]Divide by \(A_0\):
\[ \frac12=e^{-kH}. \]Taking natural logarithms gives
\[ -\ln2=-kH. \]Therefore,
\[ \boxed{H=\frac{\ln2}{k}}. \]Example: Continuous Half-Life
Suppose
\[ A(t)=A_0e^{-0.18t}. \]Then
\[ H=\frac{\ln2}{0.18}. \]Therefore,
\[ \boxed{H\approx3.85\text{ time units}}. \]Why the Same \(\ln2/k\) Appears in Both Formulas
Continuous doubling and continuous halving both ask for a factor of \(2\) in magnitude.
Growth solves
\[ e^{kD}=2, \]while decay solves
\[ e^{-kH}=\frac12. \]Since
\[ \ln\left(\frac12\right)=-\ln2, \]both time scales reduce to \(\ln2/k\).
Equivalent Half-Life Form and \(e\)-Form
A decay model can be written
\[ A(t)=A_0\left(\frac12\right)^{t/H} \]or
\[ A(t)=A_0e^{-kt}. \]These forms describe the same process when
\[ \boxed{k=\frac{\ln2}{H}}. \]Equivalent Doubling-Time Form and \(e\)-Form
A growth model can be written
\[ A(t)=A_0\,2^{t/D} \]or
\[ A(t)=A_0e^{kt}. \]These forms are equivalent when
\[ \boxed{k=\frac{\ln2}{D}}. \]Finding the Continuous Rate from Doubling Time
Starting with
\[ D=\frac{\ln2}{k}, \]solve for \(k\):
\[ \boxed{k=\frac{\ln2}{D}}. \]Finding the Continuous Decay Constant from Half-Life
Starting with
\[ H=\frac{\ln2}{k}, \]solve for \(k\):
\[ \boxed{k=\frac{\ln2}{H}}. \]Example: Build a Half-Life Model
Suppose a substance begins at \(1200\) units and has half-life \(5\) years.
\[ \boxed{A(t)=1200\left(\frac12\right)^{t/5}}. \]After \(7\) years,
\[ A(7)=1200\left(\frac12\right)^{7/5}. \]Therefore,
\[ \boxed{A(7)\approx454.71}. \]Example: Build a Doubling-Time Model
Suppose a quantity begins at \(300\) and doubles every \(9\) hours.
\[ \boxed{Q(t)=300\,2^{t/9}}. \]After \(18\) hours,
\[ Q(18)=300\,2^2=1200. \]Therefore,
\[ \boxed{Q(18)=1200}. \]Reading a Doubling-Time Model
Consider
\[ P(t)=5000\,2^{t/12}. \]- The initial value is \(5000\).
- The doubling time is \(12\) time units.
- After \(12\) units, the value is \(10000\).
- After \(24\) units, the value is \(20000\).
Reading a Half-Life Model
Consider
\[ A(t)=960\left(\frac12\right)^{t/8}. \]- The initial amount is \(960\).
- The half-life is \(8\) time units.
- After \(8\) units, \(480\) remains.
- After \(16\) units, \(240\) remains.
- After \(24\) units, \(120\) remains.
Finding Elapsed Half-Lives
If the half-life is \(H\), then after time \(t\) the number of half-lives elapsed is
\[ \boxed{n=\frac{t}{H}}. \]The remaining fraction is
\[ \boxed{\left(\frac12\right)^n}. \]Fraction Remaining after Several Half-Lives
\[ \begin{array}{c|c|c} \text{Half-lives elapsed} & \text{Fraction remaining} & \text{Percent remaining}\\ \hline 0 & 1 & 100\%\\ 1 & \frac12 & 50\%\\ 2 & \frac14 & 25\%\\ 3 & \frac18 & 12.5\%\\ 4 & \frac1{16} & 6.25\%\\ 5 & \frac1{32} & 3.125\% \end{array} \]Percentage Gone versus Percentage Remaining
After three half-lives,
\[ \frac18=12.5\% \]remains.
Therefore, the percentage gone is
\[ 100\%-12.5\%=87.5\%. \]Half-Life Does Not Depend on the Initial Amount
In a fixed ideal exponential decay process, the half-life is determined by the decay factor, not by the starting amount.
If \(1000\) units become \(500\) in \(4\) hours, then under the same model \(200\) units become \(100\) in \(4\) hours.
Doubling Time Does Not Depend on the Initial Amount
Likewise, if a process doubles every \(7\) years, then
\[ 100\longrightarrow200 \]takes \(7\) years, and
\[ 5000\longrightarrow10000 \]also takes \(7\) years.
The Rule of 70 as an Approximation
For modest percentage growth rates, doubling time is sometimes estimated using
\[ \boxed{D\approx\frac{70}{p}}, \]where \(p\) is the growth rate expressed as a percentage.
At \(5\%\) growth,
\[ D\approx\frac{70}{5}=14. \]The exact exponential calculation gives approximately 14.21 years.
The Rule of 72
Another common mental approximation is
\[ \boxed{D\approx\frac{72}{p}}. \]The number \(72\) is convenient because it has many small divisors.
These rules are estimates, not exact identities.
Why Logarithms Appear
Doubling-time and half-life problems naturally lead to equations such as
\[ 2=b^t \]or
\[ \frac12=b^t. \]The unknown appears in the exponent.
Logarithms reverse exponentiation, allowing the exponent to be isolated.
Logarithms as a Preview
If
\[ b^x=y, \]then logarithmic notation expresses the same relationship as
\[ \log_b(y)=x. \]Thus
\[ b^D=2 \]can be read as
\[ D=\log_b2. \]A later lesson will develop this inverse relationship systematically.
Radioactive Decay
Half-life is widely used to describe radioactive decay.
If \(N_0\) is the initial number of undecayed nuclei and \(H\) is the half-life, then
\[ \boxed{N(t)=N_0\left(\frac12\right)^{t/H}}. \]The formula describes the expected large-scale behavior of a population of unstable nuclei.
Medication and Biological Clearance Models
Some simplified pharmacokinetic models use exponential decay to describe the amount of a substance remaining in a system.
\[ A(t)=A_0\left(\frac12\right)^{t/H}. \]Actual biological systems can require more detailed models, so the mathematical model should not be confused with medical guidance.
Population and Growth Models
Doubling time can summarize exponential population growth:
\[ P(t)=P_0\,2^{t/D}. \]Real populations eventually encounter changing resources and constraints, so constant-doubling-time models may only be appropriate over limited intervals.
Financial Doubling
If an investment grows by a fixed effective rate \(r\) per year, then
\[ A(t)=P(1+r)^t. \]The doubling time is
\[ \boxed{D=\frac{\ln2}{\ln(1+r)}}. \]Common Error: Subtracting Half the Original Amount Repeatedly
Suppose the original amount is \(100\).
Half-life does not mean subtract \(50\) during every interval.
\[ 100,\quad50,\quad25,\quad12.5,\ldots \]Each interval halves the current amount.
Common Error: Treating Half-Life as a Linear Rate
A half-life of \(8\) years does not describe a constant amount removed per year.
Half-life describes multiplicative exponential decay.
Common Error: Dividing Time by Two
The word half refers to halving the quantity, not halving elapsed time.
If the half-life is \(10\) years, then after \(20\) years
\[ \frac{20}{10}=2 \]half-lives have elapsed, leaving
\[ \left(\frac12\right)^2=\frac14. \]Common Error: Assuming Two Half-Lives Leave Nothing
One half-life leaves \(\frac12\).
Two half-lives leave
\[ \frac12\cdot\frac12=\frac14. \]Repeated halving approaches zero but does not reach zero after a finite number of ideal half-lives.
Common Error: Using the Initial Amount to Find the Time Scale
In
\[ A(t)=A_0b^t, \]doubling time or half-life is determined by \(b\).
The initial value cancels when the time scale is solved.
Common Error: Forgetting the Time Units
A half-life of \(6\) hours and a half-life of \(6\) years are radically different models.
The numerical time scale is incomplete without its units.
Common Error: Mixing Growth and Decay Formulas
For growth with \(b>1\),
\[ \boxed{D=\frac{\ln2}{\ln b}}. \]
For decay with \(0
Matching the target factor to the behavior avoids sign errors.
A Reliable Procedure When Doubling Time Is Given
- Identify the initial value \(A_0\).
- Identify the doubling time \(D\).
- Express \(t\) in the same units as \(D\).
- Compute the number of doubling intervals \(t/D\).
- Write \(A(t)=A_0\,2^{t/D}\).
- Evaluate the exponent before rounding.
- Interpret the result with the correct units.
A Reliable Procedure When Half-Life Is Given
- Identify the initial amount \(A_0\).
- Identify the half-life \(H\).
- Express \(t\) in the same units as \(H\).
- Compute the number of half-lives \(t/H\).
- Write \(A(t)=A_0(1/2)^{t/H}\).
- Evaluate the exponential factor.
- Interpret the remaining amount or percentage as requested.
A Reliable Procedure for Finding Doubling Time
- Identify the one-period growth factor \(b\).
- If given growth rate \(r\), compute \(b=1+r\).
- Write \(b^D=2\).
- Take logarithms.
- Solve \(D=\ln2/\ln b\).
- State the result in the input period's time units.
A Reliable Procedure for Finding Half-Life
- Identify the one-period decay factor \(b\).
- If given decay rate \(r\), compute \(b=1-r\).
- Write \(b^H=1/2\).
- Take logarithms.
- Solve \(H=\ln(1/2)/\ln b\).
- State the result with the correct time units.
Example: Full Doubling-Time Workflow
A quantity grows by \(6\%\) each year.
\[ b=1.06. \]Set
\[ 1.06^D=2. \]Take natural logarithms:
\[ D\ln(1.06)=\ln2. \]Therefore,
\[ \boxed{D=\frac{\ln2}{\ln(1.06)}\approx11.90\text{ years}}. \]Example: Full Half-Life Workflow
A quantity loses \(9\%\) each month.
\[ b=1-0.09=0.91. \]Set
\[ 0.91^H=\frac12. \]Therefore,
\[ \boxed{H=\frac{\ln(1/2)}{\ln(0.91)}\approx7.35\text{ months}}. \]Practice 1: Repeated Doubling
A quantity begins at \(250\) and doubles every \(4\) years. Write an exponential model.
Practice 2: Evaluate a Doubling-Time Model
Using
\[ P(t)=1000\,2^{t/5}, \]find \(P(15)\).
Practice 3: Write a Half-Life Model
A substance begins at \(480\) grams and has a half-life of \(6\) years. Write an exponential model.
Practice 4: Evaluate a Half-Life Model
A sample begins at \(960\) units and has half-life \(4\) hours. How much remains after \(12\) hours?
Practice 5: Fraction Remaining
What fraction and percentage of an initial quantity remain after \(4\) half-lives?
Practice 6: Find Doubling Time
A quantity grows by \(10\%\) per year. Find its doubling time.
Practice 7: Find Half-Life
A quantity decays by \(15\%\) per year. Find its half-life.
Practice 8: Continuous Doubling Time
For
\[ A(t)=A_0e^{0.07t}, \]find the doubling time.
Practice 9: Continuous Half-Life
For
\[ A(t)=A_0e^{-0.12t}, \]find the half-life.
Practice 10: Interpret the Model
Consider
\[ Q(t)=3200\left(\frac12\right)^{t/7}. \]State the initial value, half-life, amount remaining after \(7\) time units, and amount remaining after \(21\) time units.
Practice Answers
Practice 1
\[ \boxed{A(t)=250\,2^{t/4}}. \]Practice 2
\[ P(15)=1000\,2^{15/5} =1000\,2^3 =\boxed{8000}. \]Practice 3
\[ \boxed{A(t)=480\left(\frac12\right)^{t/6}}. \]Practice 4
Three half-lives have elapsed:
\[ \frac{12}{4}=3. \] \[ A(12)=960\left(\frac12\right)^3=120. \]Therefore,
\[ \boxed{120}. \]Practice 5
\[ \left(\frac12\right)^4=\frac1{16}. \]Therefore,
\[ \boxed{\frac1{16}=6.25\%} \]remains.
Practice 6
\[ D=\frac{\ln2}{\ln(1.10)}. \]Therefore,
\[ \boxed{D\approx7.27\text{ years}}. \]Practice 7
The remaining factor is
\[ 1-0.15=0.85. \]Therefore,
\[ H=\frac{\ln(1/2)}{\ln(0.85)}. \]Thus,
\[ \boxed{H\approx4.27\text{ years}}. \]Practice 8
\[ D=\frac{\ln2}{0.07}. \]Therefore,
\[ \boxed{D\approx9.90\text{ time units}}. \]Practice 9
\[ H=\frac{\ln2}{0.12}. \]Therefore,
\[ \boxed{H\approx5.78\text{ time units}}. \]Practice 10
The initial value is
\[ \boxed{3200}. \]The half-life is
\[ \boxed{7\text{ time units}}. \]After \(7\) time units,
\[ Q(7)=1600. \]After \(21\) time units, three half-lives have passed:
\[ Q(21)=3200\left(\frac12\right)^3=400. \]Therefore,
\[ \boxed{Q(21)=400}. \]What to Remember
A doubling-time model is
\[ \boxed{A(t)=A_0\,2^{t/D}}. \]A half-life model is
\[ \boxed{A(t)=A_0\left(\frac12\right)^{t/H}}. \]- \(D\) is the time required for a growing quantity to double.
- \(H\) is the time required for a decaying quantity to fall to half its current value.
- The exponent \(t/D\) counts doubling intervals.
- The exponent \(t/H\) counts half-lives.
- For \(A=A_0b^t\) with \(b>1\), \(D=\ln2/\ln b\).
- For \(A=A_0b^t\) with \(0
- For continuous growth \(A=A_0e^{kt}\), \(D=\ln2/k\).
- For continuous decay \(A=A_0e^{-kt}\), \(H=\ln2/k\).
- Doubling time and half-life depend on the exponential rate, not on the initial amount.
- Repeated doubling and halving are multiplicative, not linear.
The central idea is: a doubling time or half-life packages an exponential rate into a meaningful interval of repeated multiplication.
Where We Go Next
Doubling-time and half-life formulas rely on the algebra of exponents: products of powers, quotients, negative exponents, fractional exponents, and powers raised to powers.
The next lesson, Exponent Laws, consolidates these rules and shows why they follow from repeated multiplication. Those laws will then support the transition from exponential functions to their inverses, logarithms.