📌 Part I — Solving Equations
We use log ↔ exponential conversion and log properties to solve. There are two main equation types:
Single log/exp
Isolate → convert form → solve
Isolate → convert form → solve
Multiple logs
Condense to one side → Form 1 or Form 2
Condense to one side → Form 1 or Form 2
📐 Steps for Single Log or Exponential
1
Isolate the log or exponential expression on one side
2
Convert to the alternate form (log ↔ exponential)
3
Solve for the variable
4
Simplify and leave in exact form (unless told to round)
✏️ Example 1 — Single Exponential
a)
4ˣ − 2 = 7
4ˣ = 9
4ˣ = 9
x = log₄ 9
b)
(2)^(x+3) + 5 = 26
(2)^(x+3) = 21
(2)^(x+3) = 21
x = log₂ 21 − 3
c)
−e^x = 3 − 4e^x
3e^x = 3 → e^x = 5/2
3e^x = 3 → e^x = 5/2
x = ln(5/2)
✏️ Example 2 — Single Log
a)
2 log₃(x+3) + 4 = 10
log₃(x+3) = 3
x+3 = 27
log₃(x+3) = 3
x+3 = 27
x = 24
b)
2 + log₅(2x−1) = 4
log₅(2x−1) = 2
2x−1 = 25
log₅(2x−1) = 2
2x−1 = 25
x = 13
c)
ln(2−x)/5 = 1
ln(2−x) = 5
2−x = e⁵
ln(2−x) = 5
2−x = e⁵
x = 2 − e⁵
📐 Equations with Multiple Logarithms
Two End Forms
Form 1: log_b(f) = c
→ Convert to exponential: f = b^c
→ Convert to exponential: f = b^c
Form 2: log_b(f) = log_b(g)
→ Set equal: f = g
→ Set equal: f = g
Always check for extraneous solutions! After solving, verify that all arguments of logarithms in the original equation are positive. Discard any solutions that make a log argument ≤ 0.
✏️ Example 3 — Multiple Logs (Form 2)
a)
log(3)+log(x+4)=log(5x−2)
log[3(x+4)] = log(5x−2)
3x+12 = 5x−2
log[3(x+4)] = log(5x−2)
3x+12 = 5x−2
x = 7 ✓
b)
ln(x+1)−ln(3x−5)=ln7
ln[(x+1)/(3x−5)] = ln7
(x+1)/(3x−5) = 7
ln[(x+1)/(3x−5)] = ln7
(x+1)/(3x−5) = 7
x = 9/5 ✓
c)
log₃(x+2)+log₃(x−3)=log₃14
(x+2)(x−3)=14
x²−x−20=0 → x=5 or x=−4
(x+2)(x−3)=14
x²−x−20=0 → x=5 or x=−4
x = 5 only (−4 fails domain)
d)
ln(x)−ln(3)=2
ln(x/3)=2
x/3 = e²
ln(x/3)=2
x/3 = e²
x = 3e²
✏️ Example 4 — Check for No Solution
a)
log₄(x)−3log₄2=log₄(x+7)
log₄(x/8)=log₄(x+7)
x/8 = x+7 → x = −8
log₄(x/8)=log₄(x+7)
x/8 = x+7 → x = −8
No solution (x must be > 0)
b)
log(x)+3log(2)=log(2x−7)
log(8x)=log(2x−7)
8x = 2x−7 → x = −7/6
log(8x)=log(2x−7)
8x = 2x−7 → x = −7/6
No solution (x must be > 7/2)
✏️ Examples 5–6 — Intersection Points (Quadratic)
Ex 5
ln(x+10)−ln(x)=ln(x+3)
ln[(x+10)/x]=ln(x+3)
x+10 = x(x+3) = x²+3x
x²+2x−10=0
ln[(x+10)/x]=ln(x+3)
x+10 = x(x+3) = x²+3x
x²+2x−10=0
x = (−2+√44)/2 ≈ 2.317
(negative root rejected)
Ex 6
log₃(2x−3)+log₃(x+4)=log₃(8x+2)
(2x−3)(x+4)=8x+2
2x²−3x−14=0
(2x−7)(x+2)=0
(2x−3)(x+4)=8x+2
2x²−3x−14=0
(2x−7)(x+2)=0
x = 7/2 only (−2 fails domain)
✏️ Example 7 — Common Base (Multiple Exponentials)
a)
2^(5x−1) = 2^(2x+7)
5x−1 = 2x+7
x = 8/3
b)
8^(3x−2) = 2^(x+5)
(2³)^(3x−2) = 2^(x+5)
9x−6 = x+5
(2³)^(3x−2) = 2^(x+5)
9x−6 = x+5
x = 11/8
c)
9^(2x) = 27^(x+2)
(3²)^(2x) = (3³)^(x+2)
4x = 3x+6
(3²)^(2x) = (3³)^(x+2)
4x = 3x+6
x = 6
d)
(1/2)^(5x+7) = 4^x
2^(−5x−7) = 2^(2x)
−5x−7 = 2x
2^(−5x−7) = 2^(2x)
−5x−7 = 2x
x = −1
🔁 Part II — Finding Inverses of Exp & Log Functions
Steps to Find f ⁻¹(x)
1
Switch x and y in the equation
2
Use inverse operations to solve for y (reverse order of operations)
3
Write the final answer as f ⁻¹(x) = ...
Ex 1: Find f ⁻¹
f(x) = 3(2)^(x+4) − 15
y = 3(2)^(x+4) − 15
→ switch x↔y:
x = 3(2)^(y+4) − 15
(x+15)/3 = 2^(y+4)
log₂((x+15)/3) = y+4
y = 3(2)^(x+4) − 15
→ switch x↔y:
x = 3(2)^(y+4) − 15
(x+15)/3 = 2^(y+4)
log₂((x+15)/3) = y+4
f⁻¹(x) = log₂((x+15)/3) − 4
Ex 2: Find g ⁻¹
g(x) = −3log₅(x−4) + 6
→ switch x↔y:
x = −3log₅(y−4)+6
(6−x)/3 = log₅(y−4)
5^((6−x)/3) = y−4
→ switch x↔y:
x = −3log₅(y−4)+6
(6−x)/3 = log₅(y−4)
5^((6−x)/3) = y−4
g⁻¹(x) = 5^((6−x)/3) + 4
📊 Part II — Solving Inequalities
Strategy: Isolate the expression, use log ↔ exp conversion, then apply a sign chart at critical x-values. Check domain restrictions first!
Example 4 — Log Inequality
f(x) = log((x²−4x+7)/(x+3)) > 0 [Domain: x > −3]
log((x²−4x+7)/(x+3)) > 0
→ argument must be > 1 (since log > 0 means argument > 10⁰ = 1)
x²−4x+7 > x+3
x²−5x+4 > 0
(x−1)(x−4) > 0
log((x²−4x+7)/(x+3)) > 0
→ argument must be > 1 (since log > 0 means argument > 10⁰ = 1)
x²−4x+7 > x+3
x²−5x+4 > 0
(x−1)(x−4) > 0
Solution: −3 < x < 1 or x > 4
Example 5 — Compare Two Log Functions
g(x) = ln(x²+x−5) < h(x) = ln(x+4)
ln((x²+x−5)/(x+4)) < 0 → argument < 1
x²+x−5 < x+4 → x²−9 < 0 → (x+3)(x−3) < 0
ln((x²+x−5)/(x+4)) < 0 → argument < 1
x²+x−5 < x+4 → x²−9 < 0 → (x+3)(x−3) < 0
Solution: −3 < x < (−1+√21)/2 or (−1+√21)/2 < x < 3 (with domain constraints)
Example 6 — Exponential Inequality
k(x) = 1 − 4(2)^(x−3) ≤ −31
−4(2)^(x−3) ≤ −32
(2)^(x−3) ≥ 8 = 2³
x−3 ≥ 3
−4(2)^(x−3) ≤ −32
(2)^(x−3) ≥ 8 = 2³
x−3 ≥ 3
Solution: x ≥ 6
📝 Extended — FRQ #4 Style (No Calculator)
⚠️ AP Exam FRQ #4 — No Calculator Allowed
These problems require exact answers obtained through algebraic manipulation. Common approach: isolate, use log/exp properties, simplify carefully.
FRQ 1
f(x) = e^(x/4) / e^(3/5)
Set f(x) = e^(7/4):
(x/4) − (3/5) = 7/4
x/4 = 7/4 + 3/5 = 43/20
Set f(x) = e^(7/4):
(x/4) − (3/5) = 7/4
x/4 = 7/4 + 3/5 = 43/20
x = 43/5
FRQ 2
g(x) = 8e^(3x) − e^x
Set g(x) = 3e^x:
8e^(3x) = 4e^x
e^(3x)/e^x = 1/2
e^(2x) = 1/2
2x = ln(1/2) = −ln2
Set g(x) = 3e^x:
8e^(3x) = 4e^x
e^(3x)/e^x = 1/2
e^(2x) = 1/2
2x = ln(1/2) = −ln2
x = −ln2/2 = (1−ln2)/3
Careful: rewrite step-by-step
FRQ Tip: Leave answers in exact form (e.g., ln(5/2), not a decimal). Rewrite using properties before combining. Show all steps — partial credit is awarded for correct work even if the final answer has an error.
Card 1 of 8
Strategy
What are the steps to solve an equation with a single log or exponential?
Tap to reveal ✨
Answer
Isolate → Convert form (log↔exp) → Solve → Simplify
Step 1: get the log or exp alone. Step 2: rewrite in the other form. Step 3: solve for x. Step 4: leave exact form unless told to round.
Topic 2.13
Exponential & Log Equations
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Question 1 of 5
Solve for x (exact form): 4ˣ − 2 = 7
Isolate the exponential, then convert to log form
Question 2 of 5
Solve: 2 log₃(x + 3) + 4 = 10
Isolate the log, then convert to exponential form
Question 3 of 5
Solve: log₃(x+2) + log₃(x−3) = log₃14. Check for extraneous solutions.
Product Property, then set equal, then factor
Question 4 of 5
Solve: 9^(2x) = 27^(x+2) using common bases.
Rewrite 9 = 3² and 27 = 3³, then set exponents equal
Question 5 of 5
Find f ⁻¹(x) for f(x) = 3(2)^(x+4) − 15.
Switch x and y, isolate the exponential, convert to log
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Keep going!