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Lesson 6 — Exponential Functions

Exponential function f(x) = aˣ with a > 0, a ≠ 1. Domain ℝ, range (0,+∞). Growth and decay. Euler's number e. Exponential equations. Compound interest and continuous compounding.

Used in: 1.º ano do EM (15 anos) · Math I japonês cap. 5 · Klasse 10 alemã (Exponentialfunktion) · AP Precalculus Unit 2

f(x) = a^x, \quad a > 0,\ a \neq 1

The exponential function with base $a$ . When $a > 1$ , the function is strictly increasing; when $0 < a < 1$ , strictly decreasing. The domain is $\mathbb{R}$ and the range is $(0, +\infty)$ — it never touches zero or goes negative. The special case $a = e \approx 2{,}718$ is the natural base: the unique exponential whose derivative equals itself.

Choose your door

Rigorous notation, full derivation, hypotheses

Definition and properties

Definition

"The exponential function with base $b$ is defined by $f(x) = b^x$ , where $b > 0$ , $b \neq 1$ , and $x$ is any real number." — OpenStax College Algebra 2e §6.1

Algebraic properties

Monotonicity and injectivity

Euler's number $e$

"As $n$ increases without bound, the expression $\left(1 + \frac{1}{n}\right)^n$ approaches the irrational number $e \approx 2{,}71828$ . This number appears naturally in problems of continuous growth." — Boelkins, Active Calculus §1.6

Graph

The three most commonly used exponentials. All pass through (0, 1). 2ˣ and eˣ grow; (1/2)ˣ decays. The orange curve is the reflection of the blue one across the y-axis.

Exponential equations by equality of bases

Worked examples

Example— 4· Compound interest vs continuous compounding (advanced)

Problem: Invest R$1,000 at 8% per year for 10 years. Compare annual, monthly, and continuous compounding.

Strategy: Use $M = P(1 + r/n)^{nt}$ for discrete compounding and $M = Pe^{rt}$ for continuous.

Solution:

Annual ( $n=1$ ): $M = 1\,000 \cdot (1{,}08)^{10} \approx 2\,158{,}92$ .
Monthly ( $n=12$ ): $M = 1\,000 \cdot (1 + 0{,}08/12)^{120} \approx 2\,219{,}64$ .
Continuous: $M = 1\,000 \cdot e^{0{,}8} \approx 2\,225{,}54$ .

Check: Annual total is less than monthly total, which is less than continuous — more frequent compounding yields more. The difference between monthly and continuous is only R$5,90 (0.3%).

Source. Boelkins — Active Calculus §1.6 — CC-BY-NC-SA.

Example— 5· Carbon-14 dating (real-world modeling)

Problem: The half-life of carbon-14 is $\tau_{1/2} = 5\,730$ years. A bone contains $1/8$ of its original C-14. How old is it?

Strategy: Model $N(t) = N_0 \cdot (1/2)^{t/\tau_{1/2}}$ and solve $N(t)/N_0 = 1/8$ .

Solution:

Model: $N(t) = N_0 \cdot (1/2)^{t/5730}$ .
Set equal to the given value: $(1/2)^{t/5730} = 1/8$ .
Rewrite: $1/8 = (1/2)^3$ .
Equate exponents: $t/5\,730 = 3 \Rightarrow t = 17\,190$ years.

Check: In 3 half-lives, $1 \to 1/2 \to 1/4 \to 1/8$ . ✓ The bone is ~17,190 years old — late Pleistocene, Brazilian prehistoric fauna.

Source. OpenStax Algebra and Trigonometry 2e §6.7 — CC-BY 4.0.

Exercise list

45 exercises · 11 with worked solution (25%)

Application 26Understanding 5Modeling 8Challenge 6

Ex. 6.1UnderstandingAnswer key
Why do the values of a growing exponential function eventually surpass those of a growing linear function?
Select the correct option
Growing exponential functions eventually surpass any linear function.Linear functions always grow faster than exponential functions.Exponential and linear functions grow at the same asymptotic rate.Growing exponential functions never surpass linear functions.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 1
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A linear function grows by adding a constant at each step, while an exponential $b^x$ with $b > 1$ multiplies by $b$ at each step. Multiplicative growth always dominates additive growth for large values of $x$ .
Show step-by-step (with the why)
1. Linear: $f(x) = mx + c$ — the difference $f(x+1)-f(x) = m$ is constant.
2. Exponential: $g(x) = b^x$ — the ratio $g(x+1)/g(x) = b$ is constant.
3. For $b > 1$ , eventually $b^x > mx + c$ for all large $x$ .
Ex. 6.2Understanding
Given the formula $f(x) = a \cdot b^x$ , is it possible to determine whether the function grows or decays just by looking at the formula? Explain.
Select the correct option
Yes: if the base $b > 1$ it grows; if $0 < b < 1$ it decays.No: it is impossible to determine from the formula without computing values.Yes, but only by looking at the coefficient $a$ , not the base.No: growth and decay depend only on the domain.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 2
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In $f(x) = a \cdot b^x$ with $a > 0$ : if $b > 1$ the function grows; if $0 < b < 1$ the function decays. The base $b$ completely determines the direction.
Ex. 6.3Application
"The average annual increase in a wolf pack's population is 25 individuals." Does this represent an exponential function? Justify.
Select the correct option
No; the increase of 25 individuals per year is constant (linear).Yes; any population growth is exponential.Yes; a rate of 25 per year is multiplicative.No; it is a quadratic function because it involves animals.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 4
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An increase of 25 individuals per year is a constant addition, a characteristic of a linear function, not an exponential one. Exponential growth requires multiplication by a constant factor.
Ex. 6.4ApplicationAnswer key
"A bacterial population decreases by a factor of $\tfrac{1}{8}$ every 24 hours." Does this represent an exponential function? Justify.
Select the correct option
Yes; the population is multiplied by a constant factor every 24 hours.No; the fraction $1/8$ indicates linear growth.No; it is only exponential if the base is greater than 1.Yes, but it is growth, not decay.
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OpenStax College Algebra 2e · §6.1 · ex. 5
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Decreasing by a factor of $\frac{1}{8}$ every 24 hours is equivalent to multiplying by $\frac{1}{8}$ — a constant ratio. This is the definition of exponential decay with base $b = \frac{1}{8} < 1$ .
Ex. 6.5Application
"The value of a coin collection increased by $3{,}25\%$ per year over the last 20 years." Does this represent an exponential function? Justify.
Select the correct option
Yes; the $3{,}25\%$ annual appreciation is multiplicative.No; fixed percentages generate linear growth.Yes, but only if the period is exactly 20 years.No; coin collections never follow exponential functions.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 6
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An increase of $3{,}25\%$ per year means multiplying by $1{,}0325$ each year — a constant ratio. The model is $V(t) = V_0 \cdot (1{,}0325)^t$ , clearly exponential.
Ex. 6.6ApplicationAnswer key
The population of forest A is $A(t) = 115(1{,}025)^t$ and of forest B is $B(t) = 82(1{,}029)^t$ . Which forest grows faster?
Select the correct option
Forest B grows faster: base $1{,}029 > 1{,}025$ .Forest A grows faster: it starts with more trees.Both grow at the same asymptotic rate.Forest B grows more slowly because it starts with fewer trees.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 9
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The growth rate is determined by the base: $A(t) = 115(1{,}025)^t$ has base $1{,}025$ and $B(t) = 82(1{,}029)^t$ has base $1{,}029$ . Since $1{,}029 > 1{,}025$ , forest B grows faster.
Ex. 6.7Understanding
With $A(t) = 115(1{,}025)^t$ and $B(t) = 82(1{,}029)^t$ , which forest had more trees initially and by how many?
Select the correct option
Forest A starts larger: $115 > 82$ , a difference of 33 trees.Forest B starts larger: $82 > 115$ (misreading).Both start with the same number of trees.Forest A starts with 33 fewer trees.
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 10
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At $t=0$ : $A(0) = 115$ and $B(0) = 82$ . Forest A has $115 - 82 = 33$ more trees at the start.
Ex. 6.8Modeling
With $A(t) = 115(1{,}025)^t$ and $B(t) = 82(1{,}029)^t$ , which forest will have more trees after 20 years? By how many? (Ans: A with about 188 vs B with 145)
Select the correct option
Forest A: $\approx 189$ trees; Forest B: $\approx 148$ trees; A has more.Forest B: $\approx 189$ trees; A: $\approx 148$ trees; B has more.Forest A: $\approx 115$ trees; B: $\approx 82$ trees (no growth).Both have approximately 170 trees after 20 years.
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 11
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$A(20) = 115(1{,}025)^{20} \approx 115 \times 1{,}6386 \approx 188{,}4$ ; $B(20) = 82(1{,}029)^{20} \approx 82 \times 1{,}7623 \approx 144{,}5$ . Forest A has more trees after 20 years.
Show step-by-step (with the why)
1. Compute $(1{,}025)^{20} \approx 1{,}6386$ .
2. Compute $(1{,}029)^{20} \approx 1{,}7623$ .
3. $A(20) \approx 115 \times 1{,}6386 \approx 188$ .
4. $B(20) \approx 82 \times 1{,}7623 \approx 145$ . A has more.
Ex. 6.9Application
Does the equation $y = 300(1-t)^5$ represent exponential growth, exponential decay, or neither?
Select the correct option
Neither growth nor exponential decay; the base $(1-t)$ is not constant.Exponential growth; the base is positive.Exponential decay; the coefficient is 300.Linear growth; $t$ appears in the exponent.
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 14
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In $y = 300(1-t)^5$ , the expression $(1-t)^5$ has the variable $t$ in the BASE, not in the exponent. This is a polynomial (power) function, not an exponential.
Ex. 6.10Application
Does the equation $y = 220(1{,}06)^x$ represent exponential growth, exponential decay, or neither?
Select the correct option
Exponential growth; base $1{,}06 > 1$ .Exponential decay; base $1{,}06 < 2$ .Neither growth nor decay; the exponent is $x$ , not $t$ .Linear growth; $1{,}06$ is constant.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 15
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$y = 220(1{,}06)^x$ : base $b = 1{,}06 > 1$ , therefore exponential growth at a rate of $6\%$ per unit of $x$ .
Ex. 6.11Application
Does the equation $y = 11{,}701 \cdot (0{,}97)^t$ represent exponential growth, exponential decay, or neither?
Select the correct option
Exponential decay; base $0{,}97 < 1$ .Exponential growth; the coefficient 11,701 is large.Neither; the base $0{,}97$ is very close to 1.Linear growth; the rate of change is constant.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 17
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$y = 11701(0{,}97)^t$ : base $b = 0{,}97$ , and since $0 < 0{,}97 < 1$ , the function represents exponential decay at $3\%$ per period.
Ex. 6.12Modeling
An account is opened with an initial deposit of R$6,500 and earns interest at $3{,}6\%$ per year compounded semi-annually. How much will the account be worth after 20 years? (Ans: $\approx$ R$13,268.58)
Select the correct option
$A \approx 13{,}268{,}58$ after 20 years. $A \approx 11{,}500{,}00$ after 20 years. $A \approx 12{,}000{,}00$ after 20 years. $A \approx 14{,}000{,}00$ after 20 years.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 31
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$A = 6500\left(1+\frac{0{,}036}{2}\right)^{2 \cdot 20} = 6500(1{,}018)^{40} \approx 6500 \times 2{,}0413 \approx 13268{,}58$ .
Show step-by-step (with the why)
1. Formula: $A = P\left(1+\frac{r}{n}\right)^{nt}$ .
2. $P = 6500$ , $r = 0{,}036$ , $n = 2$ , $t = 20$ .
3. Base: $1 + 0{,}036/2 = 1{,}018$ . Exponent: $2 \times 20 = 40$ .
4. $A = 6500 \times (1{,}018)^{40} \approx 13268{,}58$ .
Ex. 6.13ChallengeAnswer key
An account is worth R$14,472.74 after earning $5{,}5\%$ per year with monthly compounding for 5 years. What was the initial deposit? (Ans: $\approx$ R$11,001)
Select the correct option
$P \approx$ R\$11.001 $P \approx$ R\$10.000 $P \approx$ R\$13.000 $P \approx$ R\$12.500
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 34
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Isolate $P$ in $A = P\left(1+\frac{r}{n}\right)^{nt}$ : $P = A / \left(1+\frac{r}{n}\right)^{nt}$ . With $A = 14472{,}74$ , $r = 0{,}055$ , $n = 12$ , $t = 5$ : $P = 14472{,}74 / (1{,}004583)^{60} \approx 14472{,}74 / 1{,}3157 \approx 11001$ .
Ex. 6.14ApplicationAnswer key
Does the equation $y = 3742 \cdot e^{0{,}75t}$ represent continuous growth, continuous decay, or neither?
Select the correct option
Continuous growth; positive exponent $0{,}75t$ .Continuous decay; base $e < 3$ .Neither growth nor decay; $e$ is a constant.Linear growth; the exponent is linear in $t$ .
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 39
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$y = 3742 \cdot e^{0{,}75t}$ : in the form $y = a e^{kt}$ with $k = 0{,}75 > 0$ . A positive exponent indicates continuous growth at a rate of $75\%$ per unit of time.
Ex. 6.15Application
Does the equation $y = 150 \cdot e^{3{,}25t}$ represent continuous growth, continuous decay, or neither?
Select the correct option
Continuous growth; positive exponent $3{,}25t$ .Continuous decay; coefficient 150 is small.Neither growth nor decay; $e$ is a constant.Continuous decay; the rate $3{,}25$ is greater than 1.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 40
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$y = 150 \cdot e^{3{,}25t}$ : form $y = a e^{kt}$ with $k = 3{,}25 > 0$ . Continuous growth at a rate of $325\%$ per unit of time.
Ex. 6.16ApplicationAnswer key
Does the equation $y = 2{,}25 \cdot e^{-2t}$ represent continuous growth, continuous decay, or neither?
Select the correct option
Continuous decay; negative exponent $-2t$ .Continuous growth; $e$ is greater than 1.Neither; the coefficient $2{,}25$ is positive.Continuous growth; the absolute value of the rate $|-2| = 2$ is large.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 41
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$y = 2{,}25 \cdot e^{-2t}$ : form $y = a e^{kt}$ with $k = -2 < 0$ . A negative exponent indicates continuous decay.
Ex. 6.17Modeling
An account is opened with R$12,000 and earns $7{,}2\%$ per year with continuous compounding. How much will it be worth after 30 years? (Ans: $\approx$ R$96,168)
Select the correct option
$A \approx$ R\$96.168,03 after 30 years. $A \approx$ R\$50.000,00 after 30 years. $A \approx$ R\$75.000,00 after 30 years. $A \approx$ R\$110.000,00 after 30 years.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 42
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Continuous compounding: $A = Pe^{rt} = 12000 \cdot e^{0{,}072 \times 30} = 12000 \cdot e^{2{,}16} \approx 12000 \times 8{,}0144 \approx 96172{,}8$ .
Show step-by-step (with the why)
1. Formula: $A = P e^{rt}$ .
2. $P = 12000$ , $r = 0{,}072$ , $t = 30$ .
3. Exponent: $rt = 0{,}072 \times 30 = 2{,}16$ .
4. $A = 12000 \cdot e^{2{,}16} \approx 12000 \times 8{,}014 \approx 96168$ .
Ex. 6.18Application
Evaluate $f(-3)$ for $f(x) = 2(5)^x$ . (Ans: $0{,}016$ )
Select the correct option
$f(-3) = 2(5)^{-3} = 2/125 = 0{,}016$ . $f(-3) = 2(-5)^3 = -250$ . $f(-3) = -6(5)^3 = -750$ . $f(-3) = 2(5^3) = 250$ .
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 44
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$f(-3) = 2 \cdot 5^{-3} = 2 \cdot \frac{1}{125} = \frac{2}{125} = 0{,}016$ .
Ex. 6.19Application
Evaluate $f(-1)$ for $f(x) = -4 \cdot 2^{x+3}$ . (Ans: $-16$ )
Select the correct option
$f(-1) = -4 \cdot 2^{(-1)+3} = -4 \cdot 2^2 = -16$ . $f(-1) = -4 \cdot 2^{-1+3} = -4 \cdot 4 = 8$ . $f(-1) = (-4 \cdot 2)^{-1+3} = (-8)^2 = 64$ . $f(-1) = -4 \cdot 2^{-1} = -2$ .
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 45
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$f(x) = -4 \cdot 2^{x+3}$ . Then $f(-1) = -4 \cdot 2^{(-1)+3} = -4 \cdot 2^2 = -4 \cdot 4 = -16$ .
Ex. 6.20Application
Evaluate $f(3)$ for $f(x) = e^x$ to 4 decimal places. (Ans: $\approx 20{,}0855$ )
Select the correct option
$f(3) = e^3 \approx 20{,}0855$ . $f(3) = 3e \approx 8{,}154$ . $f(3) = e^3 \approx 3$ . $f(3) = 3^e \approx 19{,}81$ .
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 46
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$f(3) = e^3 \approx 20{,}0855$ . Recall that $e \approx 2{,}71828$ , so $e^3 \approx 20{,}09$ .
Ex. 6.21Application
Evaluate $f(-1)$ for $f(x) = -2e^{x-1}$ to 4 decimal places. (Ans: $\approx -0{,}2707$ )
Select the correct option
$f(-1) = -2e^{(-1)-1} = -2e^{-2} \approx -0{,}2707$ . $f(-1) = -2e^{1} \approx -5{,}436$ . $f(-1) = 2e^{-2} \approx 0{,}2707$ . $f(-1) = -2e^{0} = -2$ .
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 47
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$f(-1) = -2 \cdot e^{(-1)-1} = -2 \cdot e^{-2} = -\frac{2}{e^2} \approx -\frac{2}{7{,}389} \approx -0{,}2707$ .
Show step-by-step (with the why)
1. Substitute $x = -1$ : $f(-1) = -2e^{-1-1} = -2e^{-2}$ .
2. $e^{-2} = 1/e^2 \approx 1/7{,}389 \approx 0{,}1353$ .
3. $-2 \times 0{,}1353 \approx -0{,}2707$ .
Ex. 6.22Challenge
Evaluate $f(3)$ for $f(x) = 1{,}2\,e^{2x-0{,}3}$ to 4 decimal places. (Ans: $\approx 358{,}64$ )
Select the correct option
$f(3) = 1{,}2 e^{6-0{,}3} \approx 1{,}2 e^{5{,}7} \approx 360{,}0$ . $f(3) = 1{,}2 e^{3} \approx 24{,}10$ . $f(3) = 1{,}2 e^{5{,}7} \approx 1{,}2 \times 298 = 357{,}6$ (alternative rounding). $f(3) = 1{,}2 e^{2{,}7} \approx 17{,}88$ .
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 49
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$f(3) = 1{,}2 \cdot e^{2(3)-0{,}3} = 1{,}2 \cdot e^{5{,}7} \approx 1{,}2 \times 298{,}867 \approx 358{,}64$ . (Option A is closest to the correct value.)
Ex. 6.23Modeling
The fox population in a region has an annual growth rate of $9\%$ . In 2012, 23,900 foxes were counted. What is the projected population for 2020? (Ans: $\approx 47{,}622$ )
Select the correct option
$P(2020) \approx 47{,}622$ foxes. $P(2020) \approx 35{,}000$ foxes. $P(2020) \approx 52{,}000$ foxes. $P(2020) \approx 29{,}000$ foxes.
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 61
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$P(t) = 23900 \cdot (1{,}09)^t$ . From 2012 to 2020 is $t = 8$ years. $P(8) = 23900 \cdot (1{,}09)^8 \approx 23900 \times 1{,}9926 \approx 47622$ .
Show step-by-step (with the why)
1. Model: $P(t) = 23900 \cdot (1{,}09)^t$ with $t$ starting from 2012.
2. From 2012 to 2020: $t = 8$ .
3. $(1{,}09)^8 \approx 1{,}9926$ .
4. $P(8) \approx 23900 \times 1{,}9926 \approx 47622$ .
Ex. 6.24Modeling
A scientist starts with 100 mg of a radioactive substance that decays exponentially. After 35 hours, 50 mg remain. How many milligrams will remain after 54 hours? (Ans: $\approx 35{,}4$ mg)
Select the correct option
About 35.4 mg after 54 hours.About 25 mg after 54 hours.About 50 mg after 54 hours.About 12.5 mg after 54 hours.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 62
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Half-life of 35 h: $N(t) = 100 \cdot (1/2)^{t/35}$ . At $t = 54$ : $N(54) = 100 \cdot (0{,}5)^{54/35} \approx 100 \times 0{,}354 \approx 35{,}4$ mg.
Show step-by-step (with the why)
1. Half-life $\tau = 35$ h. Model: $N(t) = 100 \cdot (0{,}5)^{t/35}$ .
2. Compute $54/35 \approx 1{,}5429$ .
3. $(0{,}5)^{1{,}5429} \approx 0{,}354$ .
4. $N(54) \approx 100 \times 0{,}354 \approx 35{,}4$ mg.
Ex. 6.25Challenge
In 1985 a house was worth R$110,000. In 2005 it was worth R$145,000. What was the annual growth rate? Assuming constant growth, what was the value in 2010? (Ans: $\approx 1{,}4\%$ ; $\approx$ R$163,834)
Select the correct option
Rate of $\approx 1{,}4\%$ per year; value in 2010 $\approx$ R\$163.834.Rate of $\approx 5\%$ per year; value in 2010 $\approx$ R\$200.000.Rate of $\approx 1{,}4\%$ per year; value in 2010 $\approx$ R\$145.000.Rate of $\approx 2\%$ per year; value in 2010 $\approx$ R\$175.000.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 63
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Model: $V(t) = 110000 \cdot b^t$ . At $t = 20$ : $110000 b^{20} = 145000 \Rightarrow b = (145/110)^{1/20} \approx 1{,}01396$ . Rate $\approx 1{,}4\%$ . In 2010 ( $t=25$ ): $V(25) \approx 110000 \times (1{,}014)^{25} \approx 163834$ .
Ex. 6.26Modeling
A car was worth R$38,000 in 2007 and R$11,000 in 2013. If the value continues to fall at the same rate, how much will it be worth in 2017? (Ans: $\approx$ R$3,218)
Select the correct option
About R\$3.218 in 2017.About R\$5.000 in 2017.About R\$1.500 in 2017.About R\$7.000 in 2017.
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 64
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Depreciation model: $V(t) = 38000 \cdot b^t$ . At $t=6$ (2013): $38000 b^6 = 11000 \Rightarrow b = (11/38)^{1/6} \approx 0{,}8474$ . In 2017 ( $t=10$ ): $V(10) \approx 38000 \times (0{,}8474)^{10} \approx 3218$ .
Ex. 6.27Challenge
Jaylen wants to save R$54,000 for a property down payment. How much must he invest in an account earning $8{,}2\%$ per year compounded daily to reach his goal in 5 years? (Ans: $\approx$ R$36,097)
Select the correct option
Must invest approximately R\$36.097.Must invest approximately R\$45.000.Must invest approximately R\$30.000.Must invest approximately R\$50.000.
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Solve online OpenStax College Algebra 2e · §6.1 · ex. 65
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Isolate $P$ : $P = \frac{A}{\left(1+r/n\right)^{nt}}$ with $A = 54000$ , $r = 0{,}082$ , $n = 365$ , $t = 5$ . $P = 54000 / (1{,}082)^5 \approx 54000 / 1{,}4955 \approx 36097$ .
Ex. 6.28Modeling
Alyssa opened a retirement account at $7{,}25\%$ per year in 2000 with an initial deposit of R$13,500. How much will it be worth in 2025 with monthly compounding? How much more with continuous compounding? (Ans: $\approx$ R$83,998; difference $\approx$ R$1,949)
Select the correct option
Monthly compounding: $\approx$ R\ $83.998; continuous:$ \approx $R\$ 85.947; difference $\approx$ R\$1.949.Monthly compounding: $\approx$ R\ $75.000; continuous:$ \approx $R\$ 80.000.Both produce exactly the same value.Monthly compounding: $\approx$ R\ $90.000; continuous:$ \approx $R\$ 85.000.
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 67
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Monthly ( $n=12$ ): $A = 13500(1+0{,}0725/12)^{300} \approx 83998$ . Continuous: $A = 13500 \cdot e^{0{,}0725 \times 25} \approx 13500 \cdot e^{1{,}8125} \approx 85947$ . Difference $\approx 1949$ .
Ex. 6.29Modeling
An account at $7\%$ per year with a deposit of R$4,000. Compare balances after 9 years with annual, quarterly, monthly, and continuous compounding. (Ans: annual $\approx$ R$7,612; continuous $\approx$ R$7,715)
Select the correct option
Annual $\approx$ R\ $7.612; quarterly$ \approx $R\$ 7.688; monthly $\approx$ R\ $7.707; continuous$ \approx $R\$ 7.715.All compounding modes produce exactly R\$7.600.Annual compounding is always greater than continuous.Annual $\approx$ R\ $9.000; continuous$ \approx $R\$ 8.000.
Select an option first
Solve online OpenStax College Algebra 2e · §6.1 · ex. 68
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Annual: $4000(1{,}07)^9 \approx 7612$ . Quarterly: $4000(1{,}0175)^{36} \approx 7688$ . Monthly: $4000(1+0{,}07/12)^{108} \approx 7707$ . Continuous: $4000 e^{0{,}63} \approx 7715$ . More frequent compounding yields a higher balance.
Ex. 6.30Understanding
What is the role of the horizontal asymptote of an exponential function in describing the end behavior of its graph?
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The horizontal asymptote indicates the value that $f(x)$ approaches as $x \to \pm\infty$ , describing end behavior.The horizontal asymptote indicates the maximum value of the function.The horizontal asymptote indicates the absolute minimum of the function.The horizontal asymptote has no relation to end behavior.
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 1
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The horizontal asymptote $y = c$ of an exponential function indicates that $f(x) \to c$ as $x \to -\infty$ (growth) or $x \to +\infty$ (decay). It describes long-run behavior.
Ex. 6.31Understanding
What is the advantage of recognizing transformations of a parent function's graph algebraically?
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It allows transformation rules to be applied from an already-known parent graph.Transformed functions have the same graph as the parent function.Algebraic transformations do not work for exponential functions.Recognizing transformations is useless when calculators are available.
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 2
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Recognizing the parent function allows you to quickly identify shifts, reflections, and stretches without recomputing everything. It saves time and reduces errors when sketching graphs of complex functions.
Ex. 6.32ApplicationAnswer key
The graph of $f(x) = 3^x$ is reflected across the $y$ -axis and stretched vertically by a factor of 4. What is the equation of $g(x)$ ? State the y-intercept, domain, and range.
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$g(x) = 4 \cdot (1/3)^x$ ; y-intercept $= 4$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $g(x) = 4 \cdot 3^x$ ; y-intercept $= 4$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $g(x) = (1/3)^x$ ; y-intercept $= 1$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $g(x) = -4 \cdot 3^{-x}$ ; y-intercept $= -4$ ; domain $\mathbb{R}$ ; range $(-\infty, 0)$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 3
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Reflection of $3^x$ across the $y$ -axis: replace $x$ with $-x$ : $3^{-x} = (1/3)^x$ . Vertical stretch by 4: $g(x) = 4(1/3)^x$ . Intercept: $g(0) = 4$ . Domain: $\mathbb{R}$ ; range: $(0,+\infty)$ .
Ex. 6.33Application
The graph of $f(x) = 10^x$ is reflected across the $x$ -axis and shifted 7 units up. What is the equation of $g(x)$ ? State the y-intercept, domain, and range.
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$g(x) = -10^x + 7$ ; y-intercept $= 6$ ; domain $\mathbb{R}$ ; range $(-\infty, 7)$ . $g(x) = 10^x + 7$ ; y-intercept $= 8$ ; domain $\mathbb{R}$ ; range $(7, +\infty)$ . $g(x) = -10^x - 7$ ; y-intercept $= -8$ ; domain $\mathbb{R}$ ; range $(-\infty, -7)$ . $g(x) = 7 - 10^{-x}$ ; y-intercept $= 6$ ; domain $\mathbb{R}$ ; range $(-\infty, 7)$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 5
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Reflection across the $x$ -axis: $-10^x$ . Shift 7 units up: $g(x) = -10^x + 7$ . Intercept: $g(0) = -1 + 7 = 6$ . Range: since $10^x > 0$ , we have $-10^x < 0$ , so $g(x) < 7$ ; range $(-\infty, 7)$ .
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1. Reflect across x-axis: multiply by $-1$ : $-10^x$ .
2. Shift 7 units up: $g(x) = -10^x + 7$ .
3. Intercept: $g(0) = -1 + 7 = 6$ .
4. Range: $10^x \in (0,+\infty)$ , so $-10^x \in (-\infty,0)$ , so $g(x) \in (-\infty, 7)$ .
Ex. 6.34Challenge
The graph of $f(x) = (1{,}68)^x$ is shifted 3 units right, stretched vertically by a factor of 2, reflected across the $x$ -axis, and shifted 3 units down. What is the equation of $g(x)$ , and what are its domain and range?
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$g(x) = -2(1{,}68)^{x-3} - 3$ ; y-intercept $\approx -3{,}210$ ; domain $\mathbb{R}$ ; range $(-\infty, -3)$ . $g(x) = 2(1{,}68)^{x-3} + 3$ ; y-intercept $\approx 3{,}210$ ; domain $\mathbb{R}$ ; range $(3, +\infty)$ . $g(x) = -2(1{,}68)^{x+3} - 3$ ; y-intercept $\approx -3{,}592$ ; domain $\mathbb{R}$ ; range $(-\infty, -3)$ . $g(x) = 2(1{,}68)^{x-3} - 3$ ; y-intercept $\approx -2{,}790$ ; domain $\mathbb{R}$ ; range $(-3, +\infty)$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 6
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Applying the transformations in order: shift 3 right ( $(1{,}68)^{x-3}$ ), stretch by 2 ( $2(1{,}68)^{x-3}$ ), reflect across x-axis ( $-2(1{,}68)^{x-3}$ ), shift 3 down ( $-2(1{,}68)^{x-3}-3$ ). Intercept: $g(0) = -2(1{,}68)^{-3}-3 \approx -2(0{,}2104)-3 \approx -3{,}421$ . Range: $(-\infty,-3)$ .
Ex. 6.35Application
Write the transformation of $f(x) = 2^x$ reflected across the $y$ -axis. Identify the horizontal asymptote, domain, and range.
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$f(x) = 2^{-x}$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $f(x) = 2^x$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $f(x) = -2^x$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(-\infty, 0)$ . $f(x) = 2^{x-1}$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 26
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Reflection of $f(x) = 2^x$ across the $y$ -axis: replace $x$ with $-x$ : $f(-x) = 2^{-x}$ . The asymptote $y=0$ is preserved; domain $\mathbb{R}$ ; range $(0,+\infty)$ .
Ex. 6.36ApplicationAnswer key
Write the transformation of $f(x) = 2^x$ shifted 3 units up. Identify the asymptote, domain, and range.
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$h(x) = 2^x + 3$ ; asymptote $y=3$ ; domain $\mathbb{R}$ ; range $(3, +\infty)$ . $h(x) = 2^{x+3}$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $h(x) = 3 \cdot 2^x$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $h(x) = 2^x - 3$ ; asymptote $y=-3$ ; domain $\mathbb{R}$ ; range $(-3, +\infty)$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 27
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Shift 3 units up: $h(x) = 2^x + 3$ . The horizontal asymptote rises from $y=0$ to $y=3$ . Domain: $\mathbb{R}$ ; range: $(3, +\infty)$ .
Ex. 6.37Application
Write the transformation of $f(x) = 2^x$ shifted 2 units down. Identify the asymptote, domain, and range.
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$f(x) = 2^x - 2$ ; asymptote $y=-2$ ; domain $\mathbb{R}$ ; range $(-2, +\infty)$ . $f(x) = 2^{x-2}$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ . $f(x) = 2^x + 2$ ; asymptote $y=2$ ; domain $\mathbb{R}$ ; range $(2, +\infty)$ . $f(x) = (1/2)^x$ ; asymptote $y=0$ ; domain $\mathbb{R}$ ; range $(0, +\infty)$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 28
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Shift 2 units down: $f(x) = 2^x - 2$ . The asymptote drops to $y=-2$ . Domain: $\mathbb{R}$ ; range: $(-2, +\infty)$ .
Ex. 6.38ApplicationAnswer key
Describe the end behavior of $f(x) = -5(4)^x - 1$ .
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As $x \to +\infty$ : $f(x) \to -\infty$ ; as $x \to -\infty$ : $f(x) \to -1$ .As $x \to +\infty$ : $f(x) \to +\infty$ ; as $x \to -\infty$ : $f(x) \to -1$ .As $x \to \pm\infty$ : $f(x) \to -1$ .As $x \to +\infty$ : $f(x) \to 0$ ; as $x \to -\infty$ : $f(x) \to -\infty$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 29
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$f(x) = -5(4)^x - 1$ : since $4^x \to +\infty$ as $x \to +\infty$ , we have $-5(4)^x \to -\infty$ , so $f \to -\infty$ . As $x \to -\infty$ , $4^x \to 0$ , so $f \to -1$ .
Ex. 6.39Application
Describe the end behavior of $f(x) = 3\!\left(\tfrac{1}{2}\right)^x - 2$ .
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As $x \to +\infty$ : $f(x) \to -2$ ; as $x \to -\infty$ : $f(x) \to +\infty$ .As $x \to +\infty$ : $f(x) \to +\infty$ ; as $x \to -\infty$ : $f(x) \to -2$ .As $x \to \pm\infty$ : $f(x) \to -2$ .As $x \to +\infty$ : $f(x) \to 0$ ; as $x \to -\infty$ : $f(x) \to +\infty$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 30
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$f(x) = 3(1/2)^x - 2$ : since $(1/2)^x \to 0$ as $x \to +\infty$ , we have $f \to -2$ . As $x \to -\infty$ , $(1/2)^x \to +\infty$ , so $f \to +\infty$ .
Ex. 6.40Application
Describe the end behavior of $f(x) = 3(4)^{-x} + 2$ .
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As $x \to +\infty$ : $f(x) \to 2$ ; as $x \to -\infty$ : $f(x) \to +\infty$ .As $x \to +\infty$ : $f(x) \to +\infty$ ; as $x \to -\infty$ : $f(x) \to 2$ .As $x \to \pm\infty$ : $f(x) \to 2$ .As $x \to +\infty$ : $f(x) \to 0$ ; as $x \to -\infty$ : $f(x) \to +\infty$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 31
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$f(x) = 3(4)^{-x} + 2$ : since $4^{-x} = (1/4)^x \to 0$ as $x \to +\infty$ , we have $f \to 2$ . As $x \to -\infty$ , $(1/4)^x \to +\infty$ , so $f \to +\infty$ .
Ex. 6.41Application
Starting from $f(x) = 4^x$ , write the function that results from shifting $f(x)$ 4 units up. What is the horizontal asymptote?
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$g(x) = 4^x + 4$ ; asymptote $y = 4$ . $g(x) = 4^{x+4}$ ; asymptote $y = 0$ . $g(x) = 4 \cdot 4^x$ ; asymptote $y = 0$ . $g(x) = 4^x - 4$ ; asymptote $y = -4$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 32
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Shift 4 units up: $g(x) = 4^x + 4$ . The horizontal asymptote rises from $y=0$ to $y=4$ .
Ex. 6.42Application
Starting from $f(x) = 4^x$ , write the function that results from shifting $f(x)$ 3 units down. What is the horizontal asymptote?
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$g(x) = 4^x - 3$ ; asymptote $y = -3$ . $g(x) = 4^{x-3}$ ; asymptote $y = 0$ . $g(x) = 4^x + 3$ ; asymptote $y = 3$ . $g(x) = (1/4)^x$ ; asymptote $y = 0$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 33
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Shift 3 units down: $g(x) = 4^x - 3$ . The asymptote drops from $y=0$ to $y=-3$ .
Ex. 6.43ApplicationAnswer key
Starting from $f(x) = 4^x$ , write the function that results from shifting $f(x)$ 2 units left.
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$g(x) = 4^{x+2}$ ; asymptote $y = 0$ . $g(x) = 4^{x-2}$ ; asymptote $y = 0$ . $g(x) = 4^x + 2$ ; asymptote $y = 2$ . $g(x) = 2 \cdot 4^x$ ; asymptote $y = 0$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 34
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Shift 2 units left: replace $x$ with $x+2$ : $g(x) = 4^{x+2}$ . Horizontal shifts do not change the horizontal asymptote $y=0$ .
Ex. 6.44Application
Starting from $f(x) = 4^x$ , write the function that results from reflecting $f(x)$ across the $x$ -axis. Identify the asymptote and range.
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$g(x) = -4^x$ ; asymptote $y = 0$ ; range $(-\infty, 0)$ . $g(x) = 4^{-x}$ ; asymptote $y = 0$ ; range $(0, +\infty)$ . $g(x) = -4^{-x}$ ; asymptote $y = 0$ ; range $(-\infty, 0)$ . $g(x) = 4^x + 1$ ; asymptote $y = 1$ ; range $(1, +\infty)$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 36
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Reflection across the $x$ -axis: $g(x) = -4^x$ . The asymptote $y=0$ is preserved; the range changes to $(-\infty, 0)$ .
Ex. 6.45ChallengeAnswer key
Evaluate $g(6)$ for $g(x) = \tfrac{1}{3}(7)^{x-2}$ . (Ans: $\approx 800{,}33$ )
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$g(6) = \tfrac{1}{3}(7)^{6-2} = \tfrac{1}{3}(7)^4 = \tfrac{2401}{3} \approx 800{,}33$ . $g(6) = \tfrac{1}{3}(7)^6 = 39{,}237$ . $g(6) = \tfrac{1}{3}(7)^4 = 57{,}17$ . $g(6) = (7)^4 = 2401$ .
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Solve online OpenStax College Algebra 2e · §6.2 · ex. 43
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$g(x) = \frac{1}{3}(7)^{x-2}$ . Then $g(6) = \frac{1}{3}(7)^{6-2} = \frac{1}{3}(7)^4 = \frac{1}{3} \cdot 2401 \approx 800{,}33$ .
Show step-by-step (with the why)
1. Substitute $x = 6$ : $g(6) = \frac{1}{3}(7)^{6-2}$ .
2. Simplify the exponent: $6-2 = 4$ .
3. Compute: $7^4 = 2401$ .
4. $g(6) = \frac{2401}{3} \approx 800{,}33$ .

Sources

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

OpenStax College Algebra 2e — Jay Abramson et al. · 2022, 2nd ed. · EN · CC-BY 4.0 · §6.1 (definition and properties), §6.2 (graphs) and §6.6 (equations). Primary source for blocks A and B.
Stitz–Zeager Precalculus — Carl Stitz, Jeff Zeager · 2013, v3 · EN · CC-BY-NC-SA · §6.1 (exponential equations) and §6.3 (equations and inequalities).
Active Calculus 2.0 — Matt Boelkins · 2024 · EN · CC-BY-NC-SA · §1.6 (number $e$ , continuous compounding).
OpenStax Algebra and Trigonometry 2e — OpenStax · 2022 · EN · CC-BY 4.0 · §6.7 (models: interest, decay, population growth). Primary source for block C.
Lebl — Notes on Diffy Qs — Jiří Lebl · 2024, v6.6 · EN · CC-BY-SA · §1.4 (exponential as solution of first-order ODEs).
Hammack — Book of Proof — Richard Hammack · 2018, 3rd ed. · EN · CC-BY-ND · §10.2 (elementary proofs with exponents). Primary source for block D.