Exponential and Logarithmic Functions: Growth, Decay, Newton's Law, and Logistic Models

General Law of Uninhibited Growth or Decay

• The Exponential Law: Many natural phenomena follow the law that an amount $A$ varies with time $t$ according to the function:     * $A(t) = A_0 e^{kt}$

• Variables and Constants:     * $A_0$: The original amount present at time $t = 0$.     * $k \neq 0$: A constant representing the growth or decay rate.     * $e$: The base of the natural logarithm (approximately 2.718).

• Uninhibited Growth:     * Occurs when $k > 0$.     * The amount $A$ increases over time.

• Uninhibited Decay:     * Occurs when $k < 0$.     * The amount $A$ decreases over time.

Uninhibited Growth of Cells

• Cell Culture Model: In the early stages of growth, the number $N$ of cells in a culture after time $t$ is modeled by:     * $N(t) = N_0 e^{kt}$

• Parameters:     * $N_0$: The initial number of cells.     * $k$: A positive constant representing the growth rate of the cells.

Example 1: Bacterial Growth Analysis

• Model Provided: A colony of bacteria is modeled by the function $N(t) = 90 e^{0.06t}$, where $N$ is measured in grams and $t$ is measured in days.

• Determining Initial Amount:     * The initial amount $N_0$ is obtained when $t = 0$.     * $N(0) = 90 e^{0.06 \times 0} = 90 e^0 = 90 \times 1 = 90.0$     * The initial amount is $90.0\,grams$.

• Identifying Growth Rate:     * Comparing the model to $N(t) = N_0 e^{kt}$, the value of $k$ is $0.06$.     * This indicates a growth rate of $6.0\%$.

• Calculating Population at Specific Time (7 Days):     * $N(7) = 90 e^{0.06 \times 7} = 90 e^{0.42} \approx 136.97$     * The population after $7\,days$ is approximately $136.97\,grams$.

• Time to Reach a Specific Weight (250 Grams):     * Solve for $t$: $250 = 90 e^{0.06t}$     * Divide by $90$: $\frac{250}{90} = e^{0.06t}$     * Logarithmic form: $\ln\left(\frac{250}{90}\right) = 0.06t$     * $t = \frac{\ln\left(\frac{250}{90}\right)}{0.06} \approx 17.0$     * The population reaches $250\,grams$ in approximately $17.0\,days$.

• Determining Doubling Time:     * The population doubles when $N = 2 N_0$, which is $180\,grams$.     * Solve for $t$: $180 = 90 e^{0.06t}$     * Divide by $90$: $2 = e^{0.06t}$     * Logarithmic form: $\ln(2) = 0.06t$     * $t = \frac{\ln(2)}{0.06} \approx 11.55$     * The population doubles approximately every $11.55\,days$.

Example 2: Bacterial Growth and Tripling Time

• Function Formulation: If $N$ is the number of cells and $t$ is time in hours, the function follows the uninhibited growth law $N(t) = N_0 e^{kt}$.

• Finding the Growth Rate from Doubling Time:     * The number of bacteria doubles in $5\,hours$.     * Set $N(5) = 2 N_0$: $2 N_0 = N_0 e^{k \times 5}$     * $2 = e^{5k}$     * $5k = \ln(2)$     * $k = \frac{\ln(2)}{5} \approx 0.1386$

• The Growth Function: $N(t) = N_0 e^{0.1386t}$.

• Calculating Tripling Time:     * The size of the colony triples when $N(t) = 3 N_0$.     * $3 N_0 = N_0 e^{0.1386t}$     * $3 = e^{0.1386t}$     * $0.1386t = \ln(3)$     * $t = \frac{\ln(3)}{0.1386} \approx 7.925$     * It takes approximately $7.925\,hours$, which is $7\,hours$ and $56\,minutes$, to triple.

• Consecutive Doubling: If the population doubles in $5\,hours$, it will double a second time (becoming four times the initial size) in another $5\,hours$, for a total of $10\,hours$.

Uninhibited Radioactive Decay

• Decay Formula: The amount $A$ of radioactive material present at time $t$ is:     * $A(t) = A_0 e^{kt}$

• Variable Definitions:     * $A_0$: Original amount of radioactive material.     * $k$: A negative number ($k < 0$) representing the rate of decay.

Example 3: Estimating Age of Ancient Tools (Carbon-14 Dating)

• Problem Scenario: Burned wood and stone tools at an archeological dig contain $2.33\%$ of the original amount of Carbon-14. The half-life of Carbon-14 is $5730\,years$.

• Step 1: Finding the Decay Constant $k$:     * At $t = 5730$, half of the original amount remains ($A = 0.5 A_0$).     * $0.5 A_0 = A_0 e^{k \times 5730}$     * $0.5 = e^{5730k}$     * $5730k = \ln(0.5)$     * $k = \frac{\ln(0.5)}{5730} \approx -0.000121$

• Step 2: Determining the Age of the Artifact:     * The amount currently present is $2.33\%$ of original, so $A(t) = 0.0233 A_0$.     * $0.0233 A_0 = A_0 e^{-0.000121t}$     * $0.0233 = e^{-0.000121t}$     * $-0.000121t = \ln(0.0233)$     * $t = \frac{\ln(0.0233)}{-0.000121} \approx 31077$     * The tree was cut and burned approximately $31077\,years$ ago.

Newton’s Law of Cooling

• Theoretical Model: The temperature $u$ of a heated object at time $t$ is given by:     * $u(t) = T + (u_0 - T) e^{kt}$

• Variables:     * $T$: The constant temperature of the surrounding medium.     * $u_0$: The initial temperature of the heated object.     * $k$: A negative constant ($k < 0$).

Example 4: Cooling of a Heated Object

• Initial Conditions: Object heated to $90^\circ C$, room air temperature $T = 20^\circ C$. After $t = 5\,minutes$, the object temperature $u = 75^\circ C$.

• Step 1: Finding the Constant $k$:     * $u(t) = 20 + (90 - 20) e^{kt}$     * $u(t) = 20 + 70 e^{kt}$     * Using $u(5) = 75$:     * $75 = 20 + 70 e^{5k}$     * $55 = 70 e^{5k}$     * $e^{5k} = \frac{55}{70}$     * $5k = \ln\left(\frac{55}{70}\right)$     * $k = \frac{\ln\left(\frac{55}{70}\right)}{5} \approx -0.0482$

• Step 2: Calculating Time to reach $50^\circ C$:     * $50 = 20 + 70 e^{-0.0482t}$     * $30 = 70 e^{-0.0482t}$     * $e^{-0.0482t} = \frac{30}{70}$     * $-0.0482t = \ln\left(\frac{30}{70}\right)$     * $t = \frac{\ln\left(\frac{30}{70}\right)}{-0.0482} \approx 17.6$     * The temperature reaches $50^\circ C$ after approximately $17.6\,minutes$ ($17\,minutes$ and $36\,seconds$).

• Step 3: Calculating Elapsed Time to reach $35^\circ C$:     * $35 = 20 + 70 e^{-0.0482t}$     * $15 = 70 e^{-0.0482t}$     * $e^{-0.0482t} = \frac{15}{70}$     * $-0.0482t = \ln\left(\frac{15}{70}\right)$     * $t = \frac{\ln\left(\frac{15}{70}\right)}{-0.0482} \approx 32$     * The temperature reaches $35^\circ C$ after approximately $32\,minutes$.

• Observation of Long-Term Temperature Trends:     * As $t$ increases, the exponent $-0.0482t$ grows larger in the negative direction.     * The expression $e^{-0.0482t}$ approaches zero.     * Consequently, the temperature $u$ approaches $20^\circ C$, which is the room's air temperature.

Logistic Growth and Decay Models

• General Model Equation:     * $P(t) = \frac{c}{1 + a e^{-bt}}$

• Constants:     * $a, b$, and $c$ are constants where $a > 0$ and $c > 0$.

• Growth vs. Decay:     * Logistic Growth: $b > 0$.     * Logistic Decay: $b < 0$.

• Properties of Logistic Model:     * Domain: The set of all real numbers.     * Range: The interval $(0, c)$, where $c$ represents the carrying capacity.     * Intercepts: No x-intercepts; y-intercept is at $P(0) = \frac{c}{1+a}$.     * Asymptotes: Two horizontal asymptotes exist at $y = 0$ and $y = c$.     * Monotonicity: Function is increasing if $b > 0$ and decreasing if $b < 0$.     * Inflection Point: Occurs where the population reaches half of the carrying capacity ($P(t) = \frac{c}{2}$).     * Concavity: In growth functions, the graph changes from concave up to concave down at the inflection point. In decay functions, it changes from concave down to concave up.     * Continuity: The graph is smooth and continuous without gaps or corners.

Example 5: Logistic Growth of Yeast Population

• Given Model: $P(t) = \frac{750}{1 + 74 e^{-0.530t}}$, representing yeast population in grams after $t\,hours$.

• Carrying Capacity and Growth Rate:     * Carrying capacity $c = 750\,g$.     * Growth rate $b = 0.530$ per hour.

• Initial Population:     * $P(0) = \frac{750}{1 + 74 e^{-0.530 \times 0}} = \frac{750}{1 + 74} = \frac{750}{75} = 10$     * The initial population was $10\,g$ of yeast.

• Population after 5 Hours:     * $P(5) = \frac{750}{1 + 74 e^{-0.530 \times 5}} = \frac{750}{1 + 74 e^{-2.65}} \approx 118$     * After $5\,hours$, there is approximately $118\,g$ of yeast.

• Time to reach 250 Grams:     * $250 = \frac{750}{1 + 74 e^{-0.530t}}$     * $1 + 74 e^{-0.530t} = \frac{750}{250} = 3$     * $74 e^{-0.530t} = 2$     * $e^{-0.530t} = \frac{2}{74}$     * $-0.530t = \ln\left(\frac{2}{74}\right)$     * $t = \frac{\ln\left(\frac{2}{74}\right)}{-0.530} \approx 6.86$     * It takes approximately $6.86\,hours$ ($6\,hours$ and $52\,minutes$) to reach $250\,g$.

• Time to reach One-Half Carrying Capacity:     * One-half capacity is $375\,g$.     * Using a graphing utility to intersect $Y_1 = P(t)$ and $Y_2 = 375$:     * $t \approx 8.18\,hours$ ($8\,hours$ and $11\,minutes$).

Example 6: EFISCEN Wood Product Model

• Background: The European Forest Institute (EFI) classifies wood products by life-span:     * Short: $1\,year$.     * Medium short: $4\,years$.     * Medium long: $1\,years$.     * Long: $50\,years$ (used in building industry).

• Model for Long Life-Span Products: The percentage remaining wood products after $t\,years$ is modeled by:     * $P(t) = \frac{100.3952}{1 + 0.0316 e^{0.0581t}}$

• Decay Rate: The decay rate identified in the denominator exponent is $0.0581$.

• Remaining Wood Products after 15 Years:     * $P(15) = \frac{100.3952}{1 + 0.0316 e^{0.0581 \times 15}} \approx 93.34$     * $93.34\%$ of long-life-span wood products remain after $15\,years$.

• Time to Reach 40% Remaining:     * Solve for $t$: $40 = \frac{100.3952}{1 + 0.0316 e^{0.0581t}}$     * $1 + 0.0316 e^{0.0581t} = \frac{100.3952}{40} = 2.50988$     * $0.0316 e^{0.0581t} = 1.50988$     * $e^{0.0581t} = \frac{1.50988}{0.0316} \approx 47.781$     * $0.0581t = \ln(47.781)$     * $t = \frac{\ln(47.781)}{0.0581} \approx 66.6$     * It will take approximately $66.6\,years$ for the percentage to reach $40\%$.

• Model Validity (Numerator): The numerator $100.3952$ is considered reasonable because the theoretical maximum percentage of remaining wood products is $100\%$.