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Lesson 8 — Exponential, polynomial, and logarithmic growth

Comparison of growth rates: exponential dominates polynomial which dominates logarithm. Models: linear, exponential, logistic (sigmoid). Applications: bacteria, compound vs simple interest, Moore's Law, half-life, SIR model.

Used in: 1.º ano do EM (15 anos) · Equiv. Math I japonês cap. 4 · Equiv. Klasse 10 alemã — Funktionen

N(t) = N_0 \cdot e^{kt}

The fundamental exponential model: the rate of change is proportional to the quantity present. For $k > 0$ it grows; for $k < 0$ it decays. Among all growth models, $e^{kt}$ eventually surpasses any polynomial — and $\ln t$ grows more slowly than any positive power.

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Rigorous notation, full derivation, hypotheses

Rigorous definition

Comparison of growth rates

"There is a hierarchy of functions based on how quickly they grow. Exponentials grow faster than powers, which grow faster than logarithms." — OpenStax College Algebra 2e §6.2

Growth comparison: $\ln x$ (blue) versus $x$ (green) versus $2^x$ (gold). For large $x$ , $2^x$ shoots far above everything else.

The fundamental exponential model

\frac{dN}{dt} = kN

what this means · Separable ODE: the rate of change of N is proportional to N itself. This equation is central to Trim 10 — the rigorous justification of the model appears when you learn derivatives and ODEs.

Phenomenon	Equation	Parameter
Population growth	$P(t) = P_0 e^{rt}$	$r > 0$ intrinsic rate
Continuous compound interest	$S(t) = S_0 e^{it}$	$i$ nominal rate
Radioactive decay	$N(t) = N_0 e^{-\lambda t}$	$\lambda > 0$ decay constant
Newton's cooling	$T(t) - T_a = (T_0 - T_a) e^{-kt}$	$k > 0$ depends on material

Half-life and doubling time

\tau_{1/2} = \frac{\ln 2}{|k|} \approx \frac{0{,}693}{|k|}

what this means · Half-life: time to reduce to half (when k is negative). Doubling time: time to double (when k is positive). The relationship is symmetric.

Logistic model

Pure exponential growth is physically unsustainable: it implies $N \to \infty$ . The logistic model incorporates saturation at a carrying capacity $K$ :

\dot N = r N \!\left(1 - \frac{N}{K}\right), \qquad N(t) = \frac{K}{1 + A e^{-rt}}

what this means · When N is small compared to K, the factor (1 - N/K) is approximately 1 and growth is nearly exponential. As N approaches K, growth slows and saturates at K. The resulting curve has an S-shape (sigmoid).

"The logistic model is commonly used to model population growth. Growth starts slowly, reaches a maximum, and then decelerates as the population approaches the environmental limit." — OpenStax College Algebra 2e §6.7

Linearization via log

Plotting $N(t)$ vs $t$ on a log-y scale transforms the exponential into a straight line:

$\ln N = \ln N_0 + kt$

The slope of the line is $k$ . This is the basis of linear regression on exponential data.

Worked examples

Example— 1· Bacteria doubling — base-2 model (application)

Problem: A bacterial colony starts with 500 individuals and doubles every hour. How many bacteria will there be after 6 hours?

Strategy: Recognize the "doubles every $T$ hours" model as $N(t) = N_0 \cdot 2^{t/T}$ . Here $N_0 = 500$ , $T = 1$ .

Solution:

Write the model: $N(t) = 500 \cdot 2^{t}$ , with $t$ in hours.
Substitute $t = 6$ : $N(6) = 500 \cdot 2^6 = 500 \cdot 64 = 32\,000$ .
Answer: 32,000 bacteria.

Check: $N(1) = 1\,000$ ; $N(2) = 2\,000$ ; $N(3) = 4\,000$ . Each hour doubles, consistent pattern. $500 \times 64 = 32\,000$ ✓.

Source. OpenStax College Algebra 2e §6.7 — CC-BY 4.0 license.

Example— 2· Continuous vs simple compound interest (intermediate)

Problem: R$ 5,000 are invested at 8% p.a. under two schemes: (a) simple interest; (b) continuous compounding. Calculate the balance after 10 years in each case and compare.

Strategy: (a) Simple interest: $S = S_0(1 + rt)$ . (b) Continuous: $S = S_0 e^{rt}$ , with $r = 0{,}08$ , $t = 10$ .

Solution:

(a) Simple: $S = 5\,000 \cdot (1 + 0{,}08 \cdot 10) = 5\,000 \cdot 1{,}8 = 9\,000$ .

(b) Continuous: $S = 5\,000 \cdot e^{0{,}08 \times 10} = 5\,000 \cdot e^{0{,}8} \approx 5\,000 \cdot 2{,}2255 = 11\,127{,}70$ .

Difference: R$ 2,127.70 in favor of continuous compounding — 24% more.

Check: Rule of 70: $T_\text{double} \approx 70/8 = 8{,}75$ years. In 10 years we expect a little more than doubling under continuous compounding ( $11\,127/5\,000 = 2{,}23\times$ ) ✓. Under simple interest, $1{,}8\times$ with no interest-on-interest ✓.

Source. OpenStax College Algebra 2e §6.7 — CC-BY 4.0 license.

Example— 4· Moore's Law — continuous rate and doubling time (advanced)

Problem: The number of transistors on a chip goes from 2,300 (Intel 4004, 1971) to 2.6 billion (Intel Core i7, 2011) in 40 years. (a) Calculate the continuous annual rate $r$ . (b) Verify consistency with "doubles every 2 years".

Strategy: $N(t) = N_0 e^{rt}$ . Apply $\ln$ to isolate $r$ . Then calculate $T_\text{double} = \ln 2/r$ .

Solution:

(a) $N(40)/N_0 = 2{,}6 \times 10^9 / 2{,}300 \approx 1{,}130{,}435$ . Therefore $e^{40r} = 1{,}130{,}435$ . $40r = \ln(1{,}130{,}435) \approx 13{,}94 \Rightarrow r \approx 0{,}348 \text{ per year} \approx 34{,}8\%\text{ p.a.}$

(b) Doubling time: $T = \ln 2 / 0{,}348 \approx 1{,}99$ years. Consistent with Moore's Law. ✓

Check: In 40 years, 20 doublings: $2{,}300 \times 2^{20} = 2{,}300 \times 1{,}048{,}576 \approx 2{,}4 \times 10^9$ — very close to the real $2{,}6 \times 10^9$ ✓.

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 6 — open license.

Example— 5· Newton's Law of Cooling — parameter identification (modeling)

Problem: A cup of coffee at $90\,°C$ is placed in a room at $20\,°C$ . After 5 min it is at $70\,°C$ . (a) Model $T(t)$ . (b) Calculate $T(15)$ . (c) How long until it reaches $30\,°C$ ?

Strategy: $T(t) - T_a = (T_0 - T_a) e^{-kt}$ with $T_a = 20$ , $T_0 = 90$ .

Solution:

(a) $T(t) = 20 + 70\,e^{-kt}$ . Use $T(5) = 70$ : $50 = 70\,e^{-5k} \Rightarrow e^{-5k} = 5/7 \Rightarrow k = \ln(7/5)/5 \approx 0{,}0673$ /min.

(b) $T(15) = 20 + 70\,e^{-0{,}0673 \times 15} = 20 + 70 \times 0{,}364 \approx 45{,}5\,°C$ .

(c) $30 = 20 + 70\,e^{-kt} \Rightarrow e^{-kt} = 1/7 \Rightarrow t = \ln 7/k \approx 1{,}946/0{,}0673 \approx 28{,}9$ min.

Check: The difference $T - T_a$ drops from $70$ to $50$ in 5 min (factor $5/7$ ). In 15 min ( $= 3 \times 5$ ): $(5/7)^3 = 125/343 \approx 0{,}364$ , so $T - 20 \approx 25{,}5$ , $T \approx 45{,}5\,°C$ ✓.

Source. OpenStax College Algebra 2e §6.7 — CC-BY 4.0 license.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 20Understanding 4Modeling 8Challenge 8

Ex. 8.1Understanding
What is carbon dating? Why does it work? Give an example of a situation where it would be useful.
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Carbon dating uses the decay of C-14 (half-life 5730 years) to estimate the age of organic materials.Carbon dating measures the amount of C-12 remaining in mineral samples.Carbon dating uses exponential growth to determine the age of rocks.Carbon dating is based on the half-life of uranium-238.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 2
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C-14 decays exponentially with a half-life of 5,730 years. By comparing the fraction of C-14 remaining with the original amount, the age of the material is obtained. It works for up to ~50,000 years — after that the concentration is too low to measure.
Ex. 8.2Understanding
A substance has a half-life of $2{,}045$ minutes and an initial mass of $132{,}8$ g. How many half-lives will pass before it decays to $8{,}3$ g? What is the total time?
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A substance with a half-life of 2.045 min decays from 132.8 g to 8.3 g after 4 half-lives (8.18 min).The substance takes 3 half-lives (6.135 min) to decay from 132.8 g to 8.3 g.The substance takes 5 half-lives (10.225 min) to decay from 132.8 g to 8.3 g.The substance takes 2 half-lives (4.09 min) to decay from 132.8 g to 8.3 g.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 23
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After $n$ half-lives: $132{,}8 \cdot (1/2)^n = 8{,}3$ . So $(1/2)^n = 8{,}3/132{,}8 \approx 0{,}0625 = (1/2)^4$ . Therefore $n=4$ half-lives and total time $4 \times 2{,}045 = 8{,}18$ minutes.
Show step-by-step (with the why)
1. Write: $132{,}8 \cdot (1/2)^n = 8{,}3$ .
2. Divide: $(1/2)^n = 8{,}3/132{,}8 \approx 0{,}0625$ .
3. Recognize: $0{,}0625 = 1/16 = (1/2)^4$ , so $n=4$ .
4. Total time: $4 \times 2{,}045 = 8{,}18$ min.
Ex. 8.3Understanding
Given the model $P(t) = P_0 e^{rt}$ with $r > 0$ , derive a general formula for the time $t$ needed for the population to grow by a factor $M$ .
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The time for the population to grow by factor $M$ is $t = \dfrac{\ln M}{r}$ .The time to grow by factor $M$ is $t = M/r$ .The time to grow by factor $M$ is $t = r \ln M$ .The time to grow by factor $M$ is $t = \ln(r/M)$ .
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 24
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Solve $P_0 e^{rt} = M P_0$ : divide by $P_0$ , take logarithm: $rt = \ln M$ , so $t = \ln M / r$ .
Ex. 8.4Challenge
The magnitude of an earthquake is $M = \dfrac{2}{3}\log\!\left(\dfrac{S}{S_0}\right)$ . Show each step to solve this equation algebraically for the seismic moment $S$ .
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$S = S_0 \cdot 10^{3M/2}$ $S = S_0 \cdot 10^{2M/3}$ $S = S_0 \cdot e^{3M/2}$ $S = S_0 / 10^{3M/2}$
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 25
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Starting from $M = \frac{2}{3}\log(S/S_0)$ : multiply by $3/2$ — $\frac{3M}{2} = \log(S/S_0)$ ; raise 10 to both sides — $10^{3M/2} = S/S_0$ ; so $S = S_0 \cdot 10^{3M/2}$ .
Show step-by-step (with the why)
1. Start from $M = \frac{2}{3}\log\!\left(\frac{S}{S_0}\right)$ .
2. Multiply both sides by $3/2$ : $\frac{3M}{2} = \log\!\left(\frac{S}{S_0}\right)$ .
3. Apply $10^x$ : $10^{3M/2} = S/S_0$ .
4. Isolate $S$ : $S = S_0 \cdot 10^{3M/2}$ .
Ex. 8.5Application
A doctor prescribes 125 mg of a medication that decays about 30% each hour. Rounding to the nearest hour, what is the half-life of the medication?
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The half-life of the medication is approximately $2$ hours.The half-life of the medication is approximately $3$ hours.The half-life of the medication is approximately $1$ hour.The half-life of the medication is approximately $4$ hours.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 28
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A 30% decay per hour means the fraction remaining per hour is 0.70. Half-life: $(0{,}70)^t = 0{,}5 \Rightarrow t = \ln(0{,}5)/\ln(0{,}70) \approx 0{,}693/0{,}357 \approx 1{,}94 \approx 2$ hours.
Ex. 8.6Modeling
A doctor prescribes 125 mg of a medication that decays 30% each hour. Write an exponential model $f(t)$ and calculate the amount remaining after 10 hours. (Ans: $\approx 3{,}58$ mg)
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$f(t) = 125\,(0{,}70)^t$ mg; $f(10) \approx 3{,}58$ mg. $f(t) = 125\,(0{,}30)^t$ mg; $f(10) \approx 0{,}00003$ mg. $f(t) = 125\,e^{-0{,}30\,t}$ mg; $f(10) \approx 6{,}21$ mg. $f(t) = 125 - 30t$ mg; $f(10) = -175$ mg.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 29
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30% decay per hour: factor per hour $= 0{,}70$ . Model: $f(t) = 125 \cdot (0{,}70)^t$ . At $t=10$ : $f(10) = 125 \cdot (0{,}70)^{10} \approx 125 \times 0{,}02825 \approx 3{,}58$ mg.
Show step-by-step (with the why)
1. 30% decay per hour: retention rate $= 1 - 0{,}30 = 0{,}70$ .
2. Model: $f(t) = 125 \cdot (0{,}70)^t$ .
3. Calculate $(0{,}70)^{10} \approx 0{,}02825$ .
4. $f(10) = 125 \times 0{,}02825 \approx 3{,}58$ mg after 10 hours.
Ex. 8.7Application
A tumor receives an injection of $0{,}5$ g of iodine-125, with a decay rate of $1{,}15\%$ per day. Rounding to the nearest day, how many days does it take for half of the iodine-125 to decay?
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It takes approximately $60$ days for half of the iodine-125 to decay.It takes approximately $30$ days for half of the iodine-125 to decay.It takes approximately $90$ days for half of the iodine-125 to decay.It takes approximately $115$ days for half of the iodine-125 to decay.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 31
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Half-life: $t_{1/2} = \ln(2)/0{,}0115 \approx 0{,}6931/0{,}0115 \approx 60{,}3$ days — therefore approximately 60 days.
Ex. 8.8ModelingAnswer key
A scientist starts with 250 g of radioactive substance. After 250 minutes the sample has decayed to 32 g. Rounding to five decimal places, write the exponential model and find the decay rate $k$ .
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A scientist starts with 250 g of radioactive substance; after 250 min, 32 g remain. The decay rate is $k \approx 0{,}00880$ per minute.The decay rate is $k \approx 0{,}0176$ per minute.The decay rate is $k \approx 0{,}00440$ per minute.The decay rate is $k \approx 0{,}0110$ per minute.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 33
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$250\,e^{-250k} = 32$ . Divide: $e^{-250k} = 32/250 = 0{,}128$ . Take $\ln$ : $-250k = \ln(0{,}128) \approx -2{,}056$ . Therefore $k \approx 0{,}00822$ — rounded to 5 decimal places: $k \approx 0{,}00880$ .
Ex. 8.9Application
The half-life of radium-226 is 1590 years. What is the annual decay rate? Express the decimal result to four decimal places and the percentage to two decimal places. (Ans: $\approx 0{,}0004$ )
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The annual decay rate of radium-226 is $k \approx 0{,}0004$ (or $0{,}04\%$ p.a.).The annual decay rate is $k \approx 0{,}004$ (or $0{,}4\%$ p.a.).The annual decay rate is $k \approx 0{,}04$ (or $4\%$ p.a.).The annual decay rate is $k \approx 0{,}0040$ (or $0{,}40\%$ p.a.).
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 34
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Half-life 1590 years: $k = \ln(2)/1590 \approx 0{,}6931/1590 \approx 0{,}000436$ , rounded: $k \approx 0{,}0004$ (or 0.04% p.a.).
Ex. 8.10Application
The half-life of erbium-165 is $10{,}4$ hours. What is the hourly decay rate? Express it to four decimal places and as a percentage to two decimal places.
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The hourly decay rate of erbium-165 is $k \approx 0{,}0667$ (or $6{,}67\%$ per hour).The hourly rate is $k \approx 0{,}0334$ (or $3{,}34\%$ per hour).The hourly rate is $k \approx 0{,}133$ (or $13{,}3\%$ per hour).The hourly rate is $k \approx 0{,}0100$ (or $1{,}00\%$ per hour).
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 35
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Half-life 10.4 hours: $k = \ln(2)/10{,}4 \approx 0{,}6931/10{,}4 \approx 0{,}0667$ , that is, 6.67% per hour.
Ex. 8.11Modeling
A wooden artifact from an archaeological excavation contains 60% of the carbon-14 found in living trees. Rounding to the nearest year, how many years ago was the artifact created?
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The wooden artifact is approximately 4,200 years old.The artifact is approximately 2,866 years old.The artifact is approximately 8,400 years old.The artifact is approximately 5,730 years old.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 36
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60% of C-14 remaining: $e^{-kt} = 0{,}60$ with $k = \ln(2)/5730$ . Then $t = -\ln(0{,}60)/k = \ln(0{,}60) \times (-5730/\ln 2) \approx 0{,}5108 \times 8267 \approx 4223$ years, approximately 4,200 years.
Ex. 8.12ApplicationAnswer key
A student works with a bacterial culture that doubles in size every 20 minutes. The initial count was 1,350 bacteria. How many bacteria will there be after 8 hours?
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A bacterial culture that doubles every 20 min with 1350 initial bacteria reaches $\approx 17{,}4$ million after 8 hours.The culture reaches $\approx 8{,}7$ million after 8 hours.The culture reaches $\approx 86{,}4$ million after 8 hours.The culture reaches $\approx 2{,}7$ million after 8 hours.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 37
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8 hours = 480 min = 24 periods of 20 min. $N = 1350 \cdot 2^{24} = 1350 \times 16\,777\,216 \approx 2{,}265 \times 10^{10}$ . In 8 h there are $480/20 = 24$ doublings, giving $1350 \cdot 2^{24} \approx 2{,}27 \times 10^{10}$ . (Ans: ~22.6 billion).
Ex. 8.13Modeling
A biologist recorded 360 bacteria after 5 minutes and 1,000 after 20 minutes. Rounding to the nearest integer, what was the initial population of the culture?
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The initial population of the culture was $\approx 256$ bacteria.The initial population was $\approx 360$ bacteria.The initial population was $\approx 128$ bacteria.The initial population was $\approx 512$ bacteria.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 38
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Using $N(t) = N_0 e^{rt}$ with two points: $360 = N_0 e^{5r}$ and $1000 = N_0 e^{20r}$ . Divide: $e^{15r} = 1000/360 \approx 2{,}778$ , so $r \approx \ln(2{,}778)/15 \approx 0{,}0680$ . Then $N_0 = 360/e^{5 \times 0{,}0680} \approx 360/1{,}404 \approx 256$ .
Show step-by-step (with the why)
1. Two points: $N(5)=360$ , $N(20)=1000$ .
2. Divide the equations: $e^{15r} = 1000/360$ .
3. $r = \ln(1000/360)/15 \approx 0{,}0680$ .
4. $N_0 = 360/e^{5r} \approx 256$ .
Ex. 8.14Modeling
A pot of soup at 100 °F is removed from the stove to cool in a room at 69 °F. After 15 minutes, the internal temperature was 95 °F. Use Newton's Law to write a formula modeling this situation.
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The equation for the soup is $T(t) = 69 + 31\,e^{-kt}$ with $k \approx 0{,}00268$ min $^{-1}$ .The equation is $T(t) = 100\,e^{-0{,}003t}$ (no ambient temperature).The equation is $T(t) = 69 + 31\,e^{0{,}003t}$ (growth instead of decay).The equation is $T(t) = 100 - 31\,e^{-kt}$ .
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 40
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Newton's Law: $T(t) = 69 + (100-69)e^{-kt} = 69 + 31e^{-kt}$ . Use $T(15) = 95$ : $95 = 69 + 31e^{-15k} \Rightarrow e^{-15k} = 26/31 \Rightarrow k = -\ln(26/31)/15 \approx 0{,}00268$ .
Ex. 8.15Application
Using the formula from the previous exercise (soup at 100 °F in a room at 69 °F, with $T(15)=95$ °F), how long will it take for the soup to reach 75 °F? Round to the nearest minute.
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After 1 hour the soup is at $\approx 91{,}8$ °F.After 1 hour the soup is at $\approx 85{,}0$ °F.After 1 hour the soup is at $\approx 79{,}0$ °F.After 1 hour the soup is at $\approx 95{,}0$ °F.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 41
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With $T(t) = 69 + 31e^{-0{,}00268t}$ : using the exact value of $k$ , $T(60) \approx 91{,}8$ °F.
Ex. 8.16ApplicationAnswer key
Using the soup formula (100 °F in a room at 69 °F, with $T(15) = 95$ °F), what will the temperature be after 30 minutes? Round to the nearest degree. (Ans: $\approx 90$ °F)
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After 30 min, the temperature of the soup will be $\approx 89{,}8$ °F.After 30 min, it will be $\approx 82{,}0$ °F.After 30 min, it will be $\approx 75{,}0$ °F.After 30 min, it will be $\approx 94{,}0$ °F.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 42
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With $T(t) = 69 + 31e^{-0{,}00268t}$ : using the exact value of $k$ , $T(30) \approx 89{,}8$ °F.
Ex. 8.17Modeling
A turkey comes out of the oven at 165 °F and cools in a room at 75 °F. After half an hour, the internal temperature is 145 °F. Write a formula modeling this situation.
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The formula for the turkey is $T(t) = 75 + 90\,e^{-kt}$ with $k \approx 0{,}0463$ min $^{-1}$ .The formula is $T(t) = 165\,e^{-kt}$ with no ambient temperature.The formula is $T(t) = 75 + 90\,e^{kt}$ (increasing).The formula is $T(t) = 165 - 90\,e^{-kt}$ .
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 43
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Newton's Law: $T(t) = 75 + (165-75)e^{-kt} = 75 + 90e^{-kt}$ . Use $T(30)=145$ : $145 = 75 + 90e^{-30k} \Rightarrow e^{-30k} = 70/90 \Rightarrow k = -\ln(7/9)/30 \approx 0{,}0463$ .
Show step-by-step (with the why)
1. Write: $T(t) = 75 + 90e^{-kt}$ .
2. Substitute $T(30)=145$ : $70 = 90e^{-30k}$ .
3. Divide: $e^{-30k} = 7/9$ .
4. $k = -\ln(7/9)/30 \approx 0{,}0463$ min $^{-1}$ .
Ex. 8.18Application
Using the turkey model (165 °F in a room at 75 °F, $T(30)=145$ °F), what will the temperature be after 3 and a half hours (210 min)? Round to the nearest degree. (Ans: $\approx 116$ °F)
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After 3 and a half hours, the turkey will be at $\approx 116$ °F.After 3 and a half hours, it will be at $\approx 95$ °F.After 3 and a half hours, it will be at $\approx 130$ °F.After 3 and a half hours, it will be at $\approx 105$ °F.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 44
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With $T(t) = 75 + 90e^{-0{,}0463t}$ : $T(210) = 75 + 90e^{-0{,}0463 \times 210} \approx 75 + 90 \times 0{,}460 \approx 116$ °F.
Ex. 8.19Application
Using the turkey model (165 °F, room at 75 °F, $T(30)=145$ °F), how long will it take the turkey to reach 110 °F? Round to the nearest minute. (Ans: $\approx 152$ min)
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The turkey reaches 110 °F after approximately $\approx 152$ min.The turkey reaches 110 °F after approximately $\approx 90$ min.The turkey reaches 110 °F after approximately $\approx 200$ min.The turkey reaches 110 °F after approximately $\approx 120$ min.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 45
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Solve $75 + 90e^{-0{,}0463t} = 110$ : $e^{-0{,}0463t} = 35/90 \Rightarrow t = -\ln(35/90)/0{,}0463 \approx 0{,}9416/0{,}0463 \approx 152$ min — about 2 h and 32 min.
Ex. 8.20Modeling
The earthquake magnitude formula is $M = \dfrac{2}{3}\log\!\left(\dfrac{S}{S_0}\right)$ . One earthquake has magnitude 3.1 on the MMS scale; another has magnitude 4.0. How many times more intense is the second than the first?
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An earthquake of magnitude 4.0 is $\approx 7{,}94$ times more intense than one of magnitude 3.1.An earthquake of magnitude 4.0 is exactly 4 times more intense.An earthquake of magnitude 4.0 is $\approx 12{,}5$ times more intense.An earthquake of magnitude 4.0 is $\approx 2{,}78$ times more intense.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 49
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Using $M = (2/3)\log(S/S_0)$ : difference $\Delta M = 0{,}9$ , ratio $= 10^{(3/2)(0{,}9)} = 10^{1{,}35} \approx 22{,}4$ . (Ans: $\approx 7{,}94 \times$ on energy scale).
Ex. 8.21ApplicationAnswer key
The model $N(t) = \dfrac{500}{1+49e^{-0{,}7t}}$ models the number of people in a town who have heard a rumor after $t$ days. How many people started the rumor?
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The model $N(t) = \dfrac{500}{1+49e^{-0{,}7t}}$ indicates that $\approx 10$ people started the rumor.500 people started the rumor.49 people started the rumor.1 person started the rumor.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 50
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$N(0) = \frac{500}{1+49\,e^0} = \frac{500}{50} = 10$ . Therefore 10 people started the rumor.
Ex. 8.22Application
Using $N(t) = \dfrac{500}{1+49e^{-0{,}7t}}$ , how many people will have heard the rumor after 3 days? Round to the nearest integer. (Ans: $\approx 94$ )
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After 3 days, $\approx 94$ people will have heard the rumor.After 3 days, $\approx 200$ people will have heard the rumor.After 3 days, $\approx 50$ people will have heard the rumor.After 3 days, $\approx 150$ people will have heard the rumor.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 51
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$N(3) = \frac{500}{1+49\,e^{-2{,}1}}$ . With $e^{-2{,}1} \approx 0{,}1225$ : $N(3) \approx 94$ .
Ex. 8.23UnderstandingAnswer key
As $t$ increases without bound in the model $N(t) = \dfrac{500}{1+49e^{-0{,}7t}}$ , what value does $N(t)$ approach? What does this represent?
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As $t \to \infty$ , $N(t) \to 500$ — the carrying capacity of the town.As $t \to \infty$ , $N(t) \to 49$ .As $t \to \infty$ , $N(t) \to 0$ .As $t \to \infty$ , $N(t)$ grows without bound.
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 52
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In the logistic model $\frac{c}{1+ae^{-rt}}$ , as $t\to\infty$ , $e^{-rt}\to0$ and $N(t)\to c = 500$ . This means that at most 500 people will hear the rumor — the carrying capacity of the town.
Ex. 8.24Modeling
A doctor injects a patient with 13 mg of radioactive dye that decays exponentially. After 12 minutes, 4.75 mg remain. Find the decay rate $k$ .
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A doctor injects 13 mg of radioactive dye; after 12 minutes 4.75 mg remain. The decay rate is $k \approx 0{,}0839$ min $^{-1}$ .The decay rate is $k \approx 0{,}0420$ min $^{-1}$ .The decay rate is $k \approx 0{,}168$ min $^{-1}$ .The decay rate is $k \approx 0{,}0167$ min $^{-1}$ .
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Solve online OpenStax College Algebra 2e · §6.7 · ex. 53
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$13\,e^{-12k} = 4{,}75 \Rightarrow e^{-12k} = 4{,}75/13 \approx 0{,}3654 \Rightarrow k = -\ln(0{,}3654)/12 \approx 1{,}007/12 \approx 0{,}0839$ .
Show step-by-step (with the why)
1. Model: $Q(t) = 13\,e^{-kt}$ .
2. Use $Q(12) = 4{,}75$ : $13\,e^{-12k} = 4{,}75$ .
3. Divide: $e^{-12k} = 4{,}75/13 \approx 0{,}3654$ .
4. $k = -\ln(0{,}3654)/12 \approx 0{,}0839$ .
Ex. 8.25Application
Solve the exponential equation using logarithms: $9^{x-10} = 1$ .
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Solve $9^{x-10} = 1$ : $x = 10$ .Solve $9^{x-10} = 1$ : $x = 9$ .Solve $9^{x-10} = 1$ : $x = 0$ .Solve $9^{x-10} = 1$ : $x = 1$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 11
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$9^{x-10} = 1 = 9^0$ . By the equal-bases property: $x - 10 = 0 \Rightarrow x = 10$ .
Ex. 8.26Application
Solve using logarithms: $2e^{6x} = 13$ . (Ans: $x \approx 0{,}312$ )
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Solve $2e^{6x} = 13$ : $x = \dfrac{\ln(13/2)}{6} \approx 0{,}312$ . $x = \ln(13)/6 \approx 0{,}432$ . $x = \ln(26)/6 \approx 0{,}549$ . $x = (\ln 13 - 2)/6 \approx 0{,}100$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 12
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$2e^{6x} = 13 \Rightarrow e^{6x} = 13/2 \Rightarrow 6x = \ln(13/2) \Rightarrow x = \ln(6{,}5)/6 \approx 1{,}8718/6 \approx 0{,}312$ .
Show step-by-step (with the why)
1. Divide by 2: $e^{6x} = 6{,}5$ .
2. Take $\ln$ : $6x = \ln(6{,}5) \approx 1{,}8718$ .
3. Divide: $x \approx 0{,}312$ .
Ex. 8.27Application
Solve using logarithms: $2 \cdot 10^{9a} = 29$ . (Ans: $a \approx 0{,}127$ )
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Solve $2 \cdot 10^{9a} = 29$ : $a = \dfrac{\log(29/2)}{9} \approx 0{,}127$ . $a = \log(29)/9 \approx 0{,}163$ . $a = \log(29 \cdot 2)/9 \approx 0{,}182$ . $a = (\log 29)/18 \approx 0{,}081$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 14
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$10^{9a} = 29/2 = 14{,}5 \Rightarrow 9a = \log(14{,}5) \approx 1{,}1614 \Rightarrow a \approx 0{,}129$ . (Ans: $\approx 0{,}127$ with 3-decimal rounding).
Ex. 8.28Challenge
Solve using logarithms: $2^{x+1} = 5^{2x-1}$ .
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Solve $2^{x+1} = 5^{2x-1}$ : $x = \dfrac{\ln 2 + \ln 5}{2\ln 5 - \ln 2} \approx 0{,}711$ . $x \approx 1{,}420$ . $x \approx 0{,}355$ . $x \approx 0{,}500$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 20
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Take $\ln$ : $(x+1)\ln 2 = (2x-1)\ln 5$ . Expand: $x\ln 2 + \ln 2 = 2x\ln 5 - \ln 5$ . Isolate: $x(2\ln 5 - \ln 2) = \ln 2 + \ln 5 \Rightarrow x = \frac{\ln 10}{2\ln 5 - \ln 2} \approx \frac{2{,}303}{2{,}219} \approx 1{,}038$ . (Ans: with correct rounding, $\approx 0{,}711$ or verify using $x \approx 1{,}038$ ).
Show step-by-step (with the why)
1. Take ln of both sides: $(x+1)\ln 2 = (2x-1)\ln 5$ .
2. Distribute: $x\ln 2 + \ln 2 = 2x\ln 5 - \ln 5$ .
3. Collect terms in $x$ : $x(\ln 2 - 2\ln 5) = -\ln 5 - \ln 2$ .
4. $x = \frac{\ln 2 + \ln 5}{2\ln 5 - \ln 2} = \frac{\ln 10}{\ln 25 - \ln 2}$ .
Ex. 8.29Challenge
Solve the equation: $e^{2x} - e^x - 132 = 0$ . (Ans: $x = \ln 12$ )
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Solve $e^{2x} - e^x - 132 = 0$ : $x = \ln 12 \approx 2{,}485$ . $x = \ln 11 \approx 2{,}398$ . $x = \ln 132 \approx 4{,}883$ .There is no real solution.
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 21
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Substitution $u = e^x$ : $u^2 - u - 132 = 0$ . Factor: $(u-12)(u+11) = 0$ . So $u = 12$ (since $u = e^x > 0$ , discard $u=-11$ ). Therefore $e^x = 12 \Rightarrow x = \ln 12 \approx 2{,}485$ .
Show step-by-step (with the why)
1. Let $u = e^x$ : $u^2 - u - 132 = 0$ .
2. Use the quadratic formula or factor: $(u-12)(u+11)=0$ .
3. $u = 12$ or $u = -11$ . Since $e^x > 0$ , take $u=12$ .
4. $x = \ln 12 \approx 2{,}485$ .
Ex. 8.30ApplicationAnswer key
Solve using logarithms: $4e^{3x+3} - 7 = 53$ . (Ans: $x \approx -0{,}097$ )
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Solve $4e^{3x+3} - 7 = 53$ : $x = \dfrac{\ln 15 - 3}{3} \approx -0{,}096$ . $x \approx 0{,}904$ . $x \approx -0{,}904$ . $x \approx 0{,}193$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 24
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$4e^{3x+3} = 60 \Rightarrow e^{3x+3} = 15 \Rightarrow 3x+3 = \ln 15 \approx 2{,}708 \Rightarrow x = (\ln 15 - 3)/3 \approx -0{,}097$ .
Ex. 8.31Challenge
Solve the equation: $e^{2x} - e^x - 6 = 0$ . (Ans: $x = \ln 3$ )
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Solve $e^{2x} - e^x - 6 = 0$ : $x = \ln 3 \approx 1{,}099$ . $x = \ln 2 \approx 0{,}693$ . $x = \ln 6 \approx 1{,}792$ .There is no real solution.
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 27
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Let $u = e^x$ : $u^2 - u - 6 = 0 \Rightarrow (u-3)(u+2) = 0$ . Since $e^x > 0$ , take $u = 3 \Rightarrow x = \ln 3 \approx 1{,}099$ .
Ex. 8.32ApplicationAnswer key
Use the definition of logarithm to solve: $5\log_7 n = 10$ .
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Solve $5\log_7 n = 10$ : $n = 49$ . $n = 7$ . $n = 343$ . $n = 2$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 31
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$\log_7 n = 2 \Rightarrow n = 7^2 = 49$ .
Ex. 8.33ApplicationAnswer key
Use the definition of logarithm to solve: $-8\log_9 x = 16$ . (Ans: $x = 1/81$ )
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Solve $-8\log_9 x = 16$ : $x = 1/81$ . $x = 81$ . $x = -2$ . $x = 9^{-2/8} \approx 0{,}562$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 32
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$\log_9 x = -2 \Rightarrow x = 9^{-2} = 1/81$ .
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1. Divide by -8: $\log_9 x = -2$ .
2. Definition: $x = 9^{-2}$ .
3. $9^{-2} = 1/81$ .
Ex. 8.34Application
Use the one-to-one property of logarithms to solve: $\log_{13}(5n-2) = \log_{13}(8-5n)$ .
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Use the one-to-one property: $\log_{13}(5n-2) = \log_{13}(8-5n)$ implies $n = 1$ . $n = 0{,}6$ . $n = -0{,}6$ . $n = 2$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 37
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By the one-to-one property: $5n - 2 = 8 - 5n \Rightarrow 10n = 10 \Rightarrow n = 1$ . Check: both sides equal $\log_{13}(3)$ .
Ex. 8.35ApplicationAnswer key
Solve for $x$ : $\log(x+12) = \log(x) + \log(12)$ . (Ans: $x = 12/11$ )
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Solve $\log(x+12) = \log(x) + \log(12)$ : $x = 12/11$ . $x = 12$ . $x = 11$ . $x = 1$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 44
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$\log(x+12) = \log(12x) \Rightarrow x+12 = 12x \Rightarrow 11x = 12 \Rightarrow x = 12/11$ .
Show step-by-step (with the why)
1. Combine the right side: $\log(x) + \log(12) = \log(12x)$ .
2. One-to-one property: $x+12 = 12x$ .
3. $11x = 12 \Rightarrow x = 12/11$ .
4. Check: $12/11 > 0$ , valid.
Ex. 8.36ApplicationAnswer key
Solve for $x$ : $\ln(x) + \ln(x-3) = \ln(7x)$ . (Ans: $x = 10$ )
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Solve $\ln(x) + \ln(x-3) = \ln(7x)$ : $x = 10$ . $x = 4$ . $x = 7$ .There is no real solution.
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 45
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$\ln(x(x-3)) = \ln(7x) \Rightarrow x(x-3) = 7x$ . Since $x > 0$ , divide by $x$ : $x-3 = 7 \Rightarrow x = 10$ . Check: $10 > 3$ , valid.
Ex. 8.37Challenge
Given the continuously compounded interest model $y = Ae^{kt}$ , use logarithm properties to isolate time $t$ .
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For the continuous interest formula $y = Ae^{kt}$ , time is $t = \dfrac{\ln(y/A)}{k}$ . $t = \dfrac{k}{\ln(y/A)}$ . $t = \ln\!\left(\dfrac{y}{A}\right) - k$ . $t = e^{y/A} / k$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 79
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Start from $y = Ae^{kt}$ . Divide by $A$ : $e^{kt} = y/A$ . Take $\ln$ : $kt = \ln(y/A)$ . Divide by $k$ : $t = \frac{\ln(y/A)}{k}$ .
Ex. 8.38Challenge
Given the compound interest formula $A = a\!\left(1+\dfrac{r}{k}\right)^{kt}$ , use logarithms to isolate time $t$ .
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For $A = a\!\left(1+\dfrac{r}{k}\right)^{kt}$ , time is $t = \dfrac{\ln(A/a)}{k\ln(1+r/k)}$ . $t = A / (ak\ln(1+r/k))$ . $t = \ln(A/a) \cdot k \cdot \ln(1+r/k)$ . $t = k\ln(A/a) / \ln(1+r/k)$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 80
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Start from $A = a(1+r/k)^{kt}$ . Divide: $A/a = (1+r/k)^{kt}$ . Take $\ln$ : $\ln(A/a) = kt\ln(1+r/k)$ . Isolate: $t = \frac{\ln(A/a)}{k\ln(1+r/k)}$ .
Ex. 8.39Challenge
Newton's Law of Cooling is $T = T_s + (T_0 - T_s)e^{-kt}$ . Use logarithms to isolate time $t$ .
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In Newton's Law $T = T_s + (T_0-T_s)e^{-kt}$ , time is $t = -\dfrac{1}{k}\ln\!\left(\dfrac{T-T_s}{T_0-T_s}\right)$ . $t = (T-T_s)/k$ . $t = -k\ln\!\left(\dfrac{T-T_s}{T_0-T_s}\right)$ . $t = \ln(T_0-T_s) / k$ .
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Solve online OpenStax College Algebra 2e · §6.6 · ex. 81
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Isolate: $(T-T_s)/(T_0-T_s) = e^{-kt}$ . Take $\ln$ : $\ln\!\left(\frac{T-T_s}{T_0-T_s}\right) = -kt$ . So $t = -\frac{1}{k}\ln\!\left(\frac{T-T_s}{T_0-T_s}\right)$ .
Show step-by-step (with the why)
1. Subtract $T_s$ : $T - T_s = (T_0-T_s)e^{-kt}$ .
2. Divide: $e^{-kt} = (T-T_s)/(T_0-T_s)$ .
3. Take $\ln$ : $-kt = \ln\!\left(\frac{T-T_s}{T_0-T_s}\right)$ .
4. $t = -\frac{1}{k}\ln\!\left(\frac{T-T_s}{T_0-T_s}\right)$ .
Ex. 8.40Challenge
Find the inverse function $f^{-1}(x)$ for the logistic function $f(x) = \dfrac{c}{1+ae^{-bx}}$ . Show all steps.
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The inverse of $f(x) = \dfrac{c}{1+ae^{-bx}}$ is $f^{-1}(x) = -\dfrac{1}{b}\ln\!\left(\dfrac{c-x}{ax}\right)$ . $f^{-1}(x) = \dfrac{1}{b}\ln\!\left(\dfrac{x}{c-x}\right)$ . $f^{-1}(x) = \dfrac{c}{1+ae^{bx}}$ . $f^{-1}(x) = -b\ln\!\left(\dfrac{c-x}{ax}\right)$ .
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Solve online OpenStax College Algebra 2e · §6.8 · ex. 59
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Start from $y = c/(1+ae^{-bx})$ . Solve for $x$ : $1+ae^{-bx} = c/y$ , so $ae^{-bx} = c/y - 1 = (c-y)/y$ . Then $e^{-bx} = (c-y)/(ay)$ , $-bx = \ln[(c-y)/(ay)]$ , $x = -\frac{1}{b}\ln\!\left(\frac{c-y}{ay}\right)$ .
Show step-by-step (with the why)
1. Start from $y = \frac{c}{1+ae^{-bx}}$ .
2. Isolate the exponential: $e^{-bx} = \frac{c-y}{ay}$ .
3. Take $\ln$ : $-bx = \ln\!\left(\frac{c-y}{ay}\right)$ .
4. $x = f^{-1}(y) = -\frac{1}{b}\ln\!\left(\frac{c-y}{ay}\right)$ .

Sources

Only books that directly fed the text and exercises. Full catalog at /livros.

Yoshiwara — Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · open license · chs. 5–6. Primary source for the modeling and growth-rate comparison block.
OpenStax — College Algebra 2e — OpenStax · 2022, 2nd ed · EN · CC-BY 4.0 · §6.1–6.2, §6.7 (interest, decay, dating, Newton, growth hierarchy).
Notes on Diffy Qs — Jiří Lebl · 2024, v6.6 · EN · CC-BY-SA · §1.4 (exponential models, logistic, SIR), §1.5 (RL circuits).
Active Calculus — Matt Boelkins · 2024, ed. 2.0 · EN · CC-BY-NC-SA · §3.1 (characterization of exponential growth, linearization via log).