Exponential and Logarithmic Functions
Stewart Precalculus — Chapter 4
Exponential and logarithmic functions are inverses of each other and appear everywhere in science, finance, and engineering. This guide walks through every concept in Chapter 4 with worked examples, log law reference cards, and step-by-step equation-solving strategies.
Practice Chapter 4 — 60+ Problems
Exponential equations, log equations, growth models, and applications — all with instant step-by-step solutions.
Chapter 4 — At a Glance
Exponential Functions
f(x) = b to the x, graphs, base e, transformations, domain and range
Logarithmic Functions
Inverse of exponential, log base b, common log, natural log, graphs
Laws of Logarithms
Product, quotient, power rules; change of base; expanding and condensing
Exponential Equations
Same-base method, taking log of both sides, one-to-one property
Logarithmic Equations
Exponential form, combining logs, checking for extraneous solutions
Exponential Growth and Decay
A = A-naught times e to the kt, half-life, doubling time, compound interest
Section 4.1 — Exponential Functions
An exponential function has the form f(x) = b to the x, where the base b is a positive constant not equal to 1, and x is the variable in the exponent. The key distinction from polynomial functions is where the variable lives: in a polynomial (like x squared) the variable is the base; in an exponential (like 2 to the x) the variable is the exponent.
Definition
f(x) = bˣ
- b is the base: must satisfy b > 0 and b ≠ 1
- x is the exponent: any real number
- Domain: all real numbers (−∞, ∞)
- Range: all positive real numbers (0, ∞)
- y-intercept: always (0, 1) because b to the 0 equals 1
Growth vs. Decay
Exponential Growth — b > 1
Example: f(x) = 2ˣ
- Increasing from left to right
- As x → −∞, f(x) → 0 (approaches asymptote)
- As x → +∞, f(x) → +∞ (grows without bound)
- Steeper the base, faster the growth
Exponential Decay — 0 < b < 1
Example: f(x) = (1/2)ˣ
- Decreasing from left to right
- As x → −∞, f(x) → +∞
- As x → +∞, f(x) → 0 (approaches asymptote)
- Closer b is to 0, steeper the decay
Key Properties of f(x) = b to the x
| Property | Value |
|---|---|
| Domain | All real numbers (−∞, ∞) |
| Range | All positive reals (0, ∞) |
| y-intercept | (0, 1) for every base b |
| Horizontal asymptote | y = 0 (the x-axis) |
| One-to-one | Yes — passes horizontal line test |
| Growth (b > 1) | Increasing — rises left to right |
| Decay (0 < b < 1) | Decreasing — falls left to right |
The Natural Base e ≈ 2.71828...
The number e is irrational (like pi) and arises naturally whenever we model continuous growth or change. It is defined as the value that (1 + 1/n) to the n approaches as n grows without bound. The function f(x) = e to the x is called the natural exponential function and is the most important exponential function in calculus.
e ≈ 2.71828 18284 59045...
e to the 0 = 1
e to the 1 ≈ 2.718
e to the 2 ≈ 7.389
e to the −1 ≈ 0.368
Transformations of Exponential Functions
f(x) = b^x + c
Vertical shift up by c units. Asymptote moves to y = c.
f(x) = 2^x + 3 has asymptote y = 3
f(x) = b^x − c
Vertical shift down by c units. Asymptote moves to y = −c.
f(x) = 2^x − 5 has asymptote y = −5
f(x) = b^(x−h)
Horizontal shift right by h units. Same asymptote y = 0.
f(x) = 2^(x−3) shifts the graph 3 units right
f(x) = −b^x
Reflection across the x-axis. Range becomes (−∞, 0).
f(x) = −3^x is always negative
f(x) = b^(−x)
Reflection across the y-axis. Growth becomes decay.
f(x) = 2^(−x) = (1/2)^x is a decay function
Section 4.2 — Logarithmic Functions
The logarithm is the inverse of the exponential. Where the exponential asks "what is b to the x?", the logarithm asks "to what power must I raise b to get x?" The notation log base b of x is read "log base b of x" and the answer is an exponent.
Definition — The Fundamental Equivalence
log_b(y) = x ⟺ bˣ = y
These two statements say exactly the same thing. Switch between them freely when solving problems. The log form answers "what is the exponent?" The exponential form answers "what is the result?"
Converting Between Log and Exponential Form
Log → Exponential
log₂(8) = 3
→ 2³ = 8 ✓
log₅(125) = 3
→ 5³ = 125 ✓
log(100) = 2
→ 10² = 100 ✓
ln(e) = 1
→ e¹ = e ✓
Exponential → Log
3⁴ = 81
→ log₃(81) = 4 ✓
10⁻² = 0.01
→ log(0.01) = −2 ✓
e^k = 7
→ ln(7) = k ✓
2^(−3) = 1/8
→ log₂(1/8) = −3 ✓
Common Logarithm — log(x)
Base 10. Written as log(x) with no base shown.
log(10) = 1
log(100) = 2
log(1000) = 3
log(1) = 0
log(0.1) = −1
log(0.001) = −3
Natural Logarithm — ln(x)
Base e ≈ 2.718. Written as ln(x).
ln(e) = 1
ln(e²) = 2
ln(1) = 0
ln(e^k) = k
e^(ln x) = x
ln(x) ≈ 2.303 · log(x)
Key Properties of f(x) = log_b(x)
| Property | Value |
|---|---|
| Domain | All positive reals (0, ∞) |
| Range | All real numbers (−∞, ∞) |
| x-intercept | (1, 0) for every base b |
| Vertical asymptote | x = 0 (the y-axis) |
| One-to-one | Yes — passes horizontal line test |
| b > 1 | Increasing — rises as x increases |
| 0 < b < 1 | Decreasing — falls as x increases |
Common Mistake: Domain of a Logarithm
You can only take the logarithm of a positive number. The expression log(−3) is undefined. The expression log(0) is also undefined (the asymptote is there). When solving log equations, always check that your answer makes the argument positive. Any solution that gives a zero or negative argument is extraneous and must be rejected.
Section 4.3 — Laws of Logarithms
The three core log laws (product, quotient, power) let you expand or condense logarithmic expressions. Expanding breaks one log into simpler pieces. Condensing combines multiple logs into one. Both skills appear heavily on exams.
Multiply inside → Add outside
e.g. ln(5x) = ln 5 + ln x
Divide inside → Subtract outside
e.g. log(x/100) = log x − log 100 = log x − 2
Exponent inside → Multiplier outside
e.g. ln(x⁵) = 5 ln x
Use this to evaluate any log on a calculator
e.g. log₅(100) = ln 100 / ln 5 ≈ 2.861
b to the 1 is b, so the answer is always 1
e.g. log₃(3) = 1, ln(e) = 1
b to the 0 is 1, so the answer is always 0
e.g. log(1) = 0, ln(1) = 0
Exp and log undo each other
e.g. e^(ln 7) = 7
Log and exp undo each other
e.g. ln(e^3x) = 3x
Expanding Logarithmic Expressions — Worked Examples
Example 1: Expand log( x² · y / z³ )
= log(x² · y) − log(z³) [quotient rule]
= log(x²) + log(y) − log(z³) [product rule]
= 2·log(x) + log(y) − 3·log(z) [power rule]
Example 2: Expand ln( e²x / sqrt(x+1) )
= ln(e²x) − ln( (x+1)^(1/2) ) [quotient rule]
= ln(e²) + ln(x) − (1/2)·ln(x+1) [product + power rule]
= 2 + ln(x) − (1/2)·ln(x+1) [ln(e²) = 2]
Condensing Logarithmic Expressions — Worked Examples
Example 1: Condense 3·log(x) + log(y) − 2·log(z)
= log(x³) + log(y) − log(z²) [power rule in reverse]
= log(x³ · y) − log(z²) [product rule in reverse]
= log( x³y / z² ) [quotient rule in reverse]
Example 2: Condense (1/2)·ln(a) − 3·ln(b) + ln(c)
= ln(a^(1/2)) − ln(b³) + ln(c)
= ln( sqrt(a) · c ) − ln(b³)
= ln( c·sqrt(a) / b³ )
Change of Base Formula
log_b(x) = log(x) / log(b) = ln(x) / ln(b)
Use this whenever your calculator only has log (base 10) and ln (base e) buttons and you need to evaluate a log with any other base.
log₇(50) = ln(50) / ln(7) = 3.912 / 1.946 ≈ 2.011
log₃(100) = log(100) / log(3) = 2 / 0.4771 ≈ 4.192
log₂(1000) = log(1000) / log(2) = 3 / 0.3010 ≈ 9.966
Section 4.4 — Solving Exponential Equations
An exponential equation has the variable in the exponent. The two main strategies are (1) match the bases so both sides have the same base and set exponents equal, or (2) take logarithms of both sides and apply the power rule to bring the variable down where you can solve for it.
Strategy 1 — Same Base (One-to-One Property)
If b to the x equals b to the y, then x equals y. Use when you can write both sides with the same base.
Example A: 4ˣ = 64
Rewrite 64 as a power of 4: 64 = 4³
4ˣ = 4³
x = 3
Example B: 8^(x+1) = 32^(2x)
Rewrite both as powers of 2: 8 = 2³, 32 = 2⁵
(2³)^(x+1) = (2⁵)^(2x)
2^(3x+3) = 2^(10x) [multiply exponents]
3x + 3 = 10x [set exponents equal]
3 = 7x
x = 3/7
Strategy 2 — Take Log of Both Sides
When bases cannot be matched, take ln or log of both sides. Apply the power rule to move the exponent in front.
Example C: 3ˣ = 50
ln(3ˣ) = ln(50) [take ln of both sides]
x · ln(3) = ln(50) [power rule]
x = ln(50) / ln(3)
x ≈ 3.912 / 1.099 ≈ 3.561
Example D: 5^(2x−1) = 200
ln(5^(2x−1)) = ln(200)
(2x − 1) · ln(5) = ln(200) [power rule]
2x − 1 = ln(200) / ln(5)
2x − 1 ≈ 5.298 / 1.609 ≈ 3.292
2x ≈ 4.292
x ≈ 2.146
Example E: e^(2x) = 15
ln(e^(2x)) = ln(15) [take ln of both sides]
2x = ln(15) [ln and e cancel]
x = ln(15) / 2 ≈ 1.354
Exam Tip — Two Ways to Write the Answer
For 3 to the x = 50, the exact answer is x = ln(50)/ln(3) = log(50)/log(3). The decimal approximation is x ≈ 3.561. Unless the problem asks you to round, leave it in exact form.
Section 4.5 — Solving Logarithmic Equations
A logarithmic equation has the variable inside a logarithm. The core strategy is to isolate the logarithm, then convert to exponential form to eliminate the log. Always check your answer for extraneous solutions — this is non-negotiable on exams.
Type 1 — Single Logarithm: Isolate and Exponentiate
Example A: log₄(x) = 3
Convert: 4³ = x [exponential form]
x = 64
Check: log₄(64) = log₄(4³) = 3 ✓
Example B: 2·ln(x) − 3 = 7
2·ln(x) = 10 [add 3]
ln(x) = 5 [divide by 2]
x = e⁵ [exponentiate both sides]
x = e⁵ ≈ 148.4
Check: x > 0, argument is positive ✓
Type 2 — Multiple Logs: Combine First, Then Exponentiate
Example C: log(x) + log(x − 3) = 1
log(x(x−3)) = 1 [product rule]
x(x−3) = 10¹ [convert: base 10]
x² − 3x = 10
x² − 3x − 10 = 0
(x − 5)(x + 2) = 0
x = 5 or x = −2
Check x = −2: log(−2) is undefined — EXTRANEOUS
Answer: x = 5
Example D: ln(x+5) − ln(x) = 2
ln((x+5)/x) = 2 [quotient rule]
(x+5)/x = e² [exponentiate]
x + 5 = x·e²
5 = x·e² − x = x(e² − 1)
x = 5 / (e² − 1)
x ≈ 5 / 6.389 ≈ 0.783
Check: x > 0 and x+5 > 0 ✓
Type 3 — Log Equals Log: Set Arguments Equal
Example E: log(2x + 1) = log(x + 8)
2x + 1 = x + 8 [one-to-one property]
x = 7
Check: log(15) = log(15) ✓ and 15 > 0 ✓
Critical Rule: Always Check for Extraneous Solutions
Substituting back is mandatory, not optional. When you combine logs and solve, you may generate solutions that make the original logarithm's argument zero or negative. These solutions are extraneous and do not satisfy the original equation. The domain of log_b(x) is x > 0, period.
Section 4.6 — Exponential Growth and Decay Models
The standard model A(t) = A-naught times e to the kt describes any quantity that grows or decays at a rate proportional to its current size. Here A-naught is the initial amount, k is the rate constant, and t is time. When k is positive the quantity grows; when k is negative it decays.
The Exponential Model
A(t) = A₀ · e^(kt)
Variables
- A₀ = initial amount (at t = 0)
- k = rate constant
- t = time
- A(t) = amount at time t
Sign of k
- k > 0: exponential growth
- k < 0: exponential decay
- k = 0: constant (no change)
Half-Life
t₁₂ = ln(2) / |k|
The half-life is the time for the quantity to decrease to half its current amount. Set A = A₀/2 and solve for t.
A₀/2 = A₀·e^(kt)
1/2 = e^(kt)
ln(1/2) = kt
−ln 2 = kt
t = ln 2 / |k|
Doubling Time
t₂ = ln(2) / k
The doubling time is the time for a growing quantity to double. Set A = 2A₀ and solve for t.
2A₀ = A₀·e^(kt)
2 = e^(kt)
ln 2 = kt
t = ln 2 / k
Full Worked Problem — Radioactive Decay
Iodine-131 decays from 80 mg to 50 mg in 4 days. Find the decay rate k, write the model, and find the amount remaining after 10 days.
Step 1 — Set up the equation
A(4) = 50, A₀ = 80
50 = 80·e^(4k)
Step 2 — Solve for k
50/80 = e^(4k)
0.625 = e^(4k)
ln(0.625) = 4k
−0.4700 = 4k
k ≈ −0.1175 (negative = decay)
Step 3 — Write the model
A(t) = 80·e^(−0.1175t)
Step 4 — Find amount at t = 10
A(10) = 80·e^(−1.175)
A(10) ≈ 80 · 0.309 ≈ 24.7 mg
Real-World Applications
Compound Interest
A = P(1 + r/n)^(nt)
P = principal, r = annual rate (decimal), n = periods per year, t = years
$2,000 at 4% compounded quarterly for 10 years: A = 2000(1 + 0.04/4)^40 = 2000(1.01)^40 ≈ $2,982
Continuous Compounding
A = Pe^(rt)
P = principal, r = annual rate (decimal), t = years, e ≈ 2.71828
$2,000 at 4% continuously for 10 years: A = 2000·e^(0.4) ≈ $2,983.65 — slightly more than quarterly
Exponential Growth / Decay
A(t) = A₀ · e^(kt)
A₀ = initial amount, k = rate constant (k > 0 growth, k < 0 decay), t = time
Bacteria: 500 cells, doubles every 3 hours. Find k: 2 = e^(3k) → k = ln2/3 ≈ 0.231. After 9 hours: A = 500·e^(0.231·9) ≈ 4,000 cells
Half-Life
A(t) = A₀ · (1/2)^(t/h)
A₀ = initial amount, h = half-life period, t = time (same units as h)
Carbon-14 half-life is 5,730 years. After 17,190 years (3 half-lives): A = A₀ · (1/2)³ = A₀/8. Only 12.5% remains.
Doubling Time
t_double = ln 2 / k
k = growth constant from A(t) = A₀·e^(kt)
Population grows at 2% per year (k = 0.02). Doubling time = ln2 / 0.02 = 0.693 / 0.02 ≈ 34.7 years
pH Scale
pH = −log[H⁺]
[H⁺] = hydrogen ion concentration in moles per liter
Coffee with [H⁺] = 1×10⁻⁵: pH = −log(10⁻⁵) = −(−5) = 5 (acidic)
Richter Scale
M = log(I/I₀)
I = intensity of earthquake, I₀ = reference intensity
A magnitude 7 quake is 10 times more intense than a magnitude 6 quake, and 1000 times more than a magnitude 4 quake
Frequently Asked Questions — Chapter 4
What is the difference between f(x) = 2ˣ and f(x) = x²?▾
The difference is where the variable lives. In f(x) = x² (a polynomial), the variable x is the base and the exponent 2 is constant. In f(x) = 2 to the x (an exponential), the base 2 is constant and the variable x is in the exponent.
This distinction has enormous consequences. The polynomial x² grows at a steady rate — it can be differentiated using the power rule. The exponential 2 to the x grows faster and faster — its rate of change is proportional to itself. At x = 10: x² = 100, but 2 to the x = 1,024. At x = 50: x² = 2,500, but 2 to the x is approximately 10 to the 15. Exponentials grow much faster than polynomials.
Why can't the base of an exponential be 1 or negative?▾
If b = 1, then 1 to the x = 1 for every value of x. The function is just the constant f(x) = 1 — a flat horizontal line, not an interesting exponential.
If b is negative, problems arise immediately. For example, (−4) to the (1/2) would require taking a square root of a negative number, which is not a real number. The function would be undefined for many x values and would oscillate between positive and negative. To keep the domain as all real numbers and the function well-defined, we require b > 0 and b ≠ 1.
How do I evaluate log₃(27) without a calculator?▾
Ask yourself: "To what power must I raise 3 to get 27?" Since 3³ = 27, the answer is 3. So log₃(27) = 3.
This works for any "nice" logarithm where the argument is a perfect power of the base. Practice recognizing: log₂(32) = 5 because 2⁵ = 32. log₅(125) = 3 because 5³ = 125. log₁₀(0.001) = −3 because 10^(−3) = 0.001. log₄(1/16) = −2 because 4^(−2) = 1/16.
The strategy: mentally convert to exponential form and ask what exponent is needed. The logarithm IS the exponent.
Why do we use e instead of 10 for growth and decay models?▾
The number e arises naturally when we model continuous growth. If you compound interest infinitely often (every instant), the compound interest formula (1 + r/n) to the nt approaches P times e to the rt as n grows without bound. Nature does not add interest in discrete chunks — bacteria multiply continuously, radioactive atoms decay continuously.
Using e as the base also has a beautiful mathematical property: the derivative of e to the x is exactly e to the x. This means the rate of change of an exponential with base e is proportional to the function itself — which is the defining property of exponential growth and decay. Using base 10 would introduce extra factors of ln(10) in the calculus. Base e keeps everything clean.
Can log(x + y) be simplified to log(x) + log(y)?▾
No — this is one of the most common errors in precalculus. The product rule says log(x · y) = log(x) + log(y). There is multiplication inside the log, not addition. log(x + y) cannot be simplified further.
Counter-example: log(10 + 90) = log(100) = 2. But log(10) + log(90) = 1 + 1.954 = 2.954. These are not equal.
Similarly, log(x − y) is NOT equal to log(x) − log(y). The quotient rule applies to log(x/y), not log(x − y). These are among the most frequently tested misconceptions in Chapter 4 exams.
How do I find the horizontal asymptote of an exponential function?▾
The horizontal asymptote of f(x) = b to the x is y = 0. The graph gets closer and closer to the x-axis but never touches it. When the function is shifted vertically, the asymptote shifts too.
Rule: for f(x) = A times b^(x − h) + k, the horizontal asymptote is y = k. The +k shifts the entire graph up by k units, which shifts the asymptote from y = 0 to y = k.
Examples: f(x) = 3^x + 7 has asymptote y = 7. f(x) = 2^(x−4) − 1 has asymptote y = −1. f(x) = −5·e^x + 10 has asymptote y = 10 (the negative sign flips the graph but does not change the asymptote).
What is the one-to-one property and when do I use it?▾
The one-to-one property says that both exponential and logarithmic functions with the same base are one-to-one — no two different inputs produce the same output.
For exponentials: if b to the u = b to the v, then u = v. You can set exponents equal when the bases match.
For logarithms: if log_b(u) = log_b(v), then u = v. You can set arguments equal when the bases match. Use this in Type 3 logarithmic equations where you have log(something) = log(something else) — just set the arguments equal and solve algebraically. Remember to still check for extraneous solutions.
Solve: 2^(x+1) = 3^(x−2) step by step.▾
The bases 2 and 3 cannot be matched, so we take ln of both sides.
ln(2^(x+1)) = ln(3^(x−2))
(x+1)·ln 2 = (x−2)·ln 3 [power rule]
x·ln 2 + ln 2 = x·ln 3 − 2·ln 3 [expand]
x·ln 2 − x·ln 3 = −2·ln 3 − ln 2 [collect x terms]
x(ln 2 − ln 3) = −2·ln 3 − ln 2
x = (−2·ln 3 − ln 2) / (ln 2 − ln 3)
x ≈ (−2·1.099 − 0.693) / (0.693 − 1.099)
x ≈ −2.891 / (−0.406) ≈ 7.12
The exact form is x = (2·ln 3 + ln 2) / (ln 3 − ln 2). Simplified using log laws: x = ln(9·2) / ln(3/2) = ln(18) / ln(1.5).
How do you find k from a half-life problem?▾
Plug the half-life into the model with A = A₀/2 and solve for k.
Example: Cesium-137 has a half-life of 30 years. Find k.
A₀/2 = A₀ · e^(k·30) [A halves in 30 years]
1/2 = e^(30k)
ln(1/2) = 30k
−ln 2 = 30k
k = −ln 2 / 30 ≈ −0.0231
The model is A(t) = A₀ · e^(−0.0231t). The negative sign confirms decay. To find the amount remaining after any time t, substitute t and evaluate.
What is continuous compounding and how does it differ from regular compounding?▾
Regular compounding adds interest at discrete intervals — monthly, quarterly, daily. The formula is A = P(1 + r/n) to the nt.
Continuous compounding is the theoretical limit as n approaches infinity — interest is added at every instant. The formula simplifies to A = P times e to the rt. This is derived using the limit definition of e.
$10,000 at 5% for 20 years:
Annual (n=1): A = 10000(1.05)²⁰ ≈ $26,533
Monthly (n=12): A = 10000(1.00417)²⁴⁰ ≈ $27,127
Daily (n=365): A = 10000(1.000137)^7300 ≈ $27,179
Continuous: A = 10000·e^(1.0) ≈ $27,183
More frequent compounding earns more interest, but the gains diminish and converge to the continuous compounding limit. The difference between daily and continuous is only a few dollars.
How do you expand log( cube-root of (x²y / z⁴) )?▾
Rewrite the cube root as a 1/3 power first, then apply all three log laws.
log( (x²y / z⁴)^(1/3) )
= (1/3) · log(x²y / z⁴) [power rule]
= (1/3) · [log(x²y) − log(z⁴)] [quotient rule]
= (1/3) · [log(x²) + log(y) − log(z⁴)] [product rule]
= (1/3) · [2·log x + log y − 4·log z] [power rule]
= (2/3)·log x + (1/3)·log y − (4/3)·log z
Strategy: always rewrite roots as fractional exponents, then apply power rule first on the outermost exponent, quotient/product rules on the fraction, and power rule again on any remaining exponents.
Chapter 4 Formula Reference Card
Exponential Functions
f(x) = bˣ (b > 0, b ≠ 1)
Domain: (−∞, ∞) Range: (0, ∞)
f(0) = 1 always
HA: y = 0
e ≈ 2.71828
A = Pe^(rt) (continuous)
A = P(1+r/n)^(nt) (periodic)
Logarithmic Functions
log_b(y) = x ⟺ bˣ = y
Domain: (0, ∞) Range: (−∞, ∞)
f(1) = 0 always
VA: x = 0
log = log₁₀ ln = log_e
Change of base: ln x / ln b
Log Laws
log(MN) = log M + log N
log(M/N) = log M − log N
log(Mⁿ) = n·log M
log_b(b) = 1
log_b(1) = 0
b^(log_b x) = x
log_b(bˣ) = x
Growth & Decay
A(t) = A₀·e^(kt)
k > 0: growth k < 0: decay
Half-life: t = ln 2 / |k|
Doubling: t = ln 2 / k
pH = −log[H⁺]
Richter: M = log(I/I₀)
Study Tips for Chapter 4
Memorize the fundamental equivalence
Every single log problem depends on: log_b(x) = y means b to the y equals x. If you own this one fact, you can convert freely between log and exponential form, which is the key to solving almost every problem in the chapter.
Know all three log laws cold
Product, quotient, and power. Drill them until they are automatic. Most Chapter 4 exam problems are just applying these laws in the right order. Also memorize: log_b(b) = 1, log_b(1) = 0, and log_b(b to the x) = x.
Always check extraneous solutions
Every time you solve a logarithmic equation, substitute your answer back into the original. Any solution that makes the argument of a log zero or negative is invalid. This step is worth points — do not skip it.
Sketch the graph to build intuition
Draw f(x) = 2 to the x and f(x) = log₂(x) on the same axes. Notice they are reflections across y = x. The exponential shoots up on the right and flattens on the left. The log shoots up on the left (from the asymptote) and grows slowly on the right. Visualizing this reduces errors.
For growth/decay: find k first
Every application problem gives you enough information to find k — usually an initial condition and one other data point. Find k before answering any question about the model. With k pinned down, every other question is just substitution.
Related Topics
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