Five strategies — matching bases, taking logarithms, natural exponential, change of base, and equations on both sides — plus compound interest, half-life, and doubling time applications with 5 fully worked examples.
Can both sides be written with the same base?
YES → Set exponents equal and solve
Is the base e?
YES → Take ln of both sides; ln(eˣ) = x cleans up immediately
Are there exponentials on both sides with different bases?
YES → Take ln of both sides, expand with power rule, collect x terms
Is the equation quadratic in bˣ? (e.g., (2ˣ)² − 5·2ˣ + 6 = 0)
YES → Substitute u = bˣ, solve the quadratic, then recover x (discard u ≤ 0)
None of the above — different bases, no matching possible
DEFAULT → Take ln of both sides, use power rule, solve for x
Power rule for logs
log(bˣ) = x · log(b)
Brings the variable exponent down as a multiplier — the key to all log-based solving
Natural log inverse
ln(eˣ) = x & e^(ln x) = x
ln and e are inverses — they cancel each other. Valid for x > 0 in the second identity
Change of base
log_b(x) = ln(x) / ln(b) = log(x) / log(b)
Convert any base to ln or log₁₀. Used to evaluate on any calculator
Solution when bases match
bˣ = bᵏ ⟺ x = k
One-to-one property: exponential functions with the same base are injective
Cross-base equation solution
bˣ = c → x = ln(c) / ln(b)
General formula for any base b > 0, b ≠ 1, and any c > 0
Each method below includes when to use it, the step-by-step procedure, and a worked example. Learn all five to handle any exponential equation you encounter.
Use when: Both sides can be written with the same base
Steps
Example
Solve: 4ˣ⁺¹ = 32
Rewrite: (2²)ˣ⁺¹ = 2⁵
Simplify: 2²⁽ˣ⁺¹⁾ = 2⁵
Equate exponents: 2(x + 1) = 5
Expand: 2x + 2 = 5
x = 3/2
Use when: Bases cannot be matched (most general method)
Steps
Example
Solve: 3ˣ = 7
Take ln of both sides: ln(3ˣ) = ln(7)
Power rule: x · ln(3) = ln(7)
Divide: x = ln(7)/ln(3)
x ≈ 1.946/1.099 ≈ 1.771
Use when: The base is e (eˣ = k)
Steps
Example
Solve: e²ˣ⁻¹ = 9
Take ln of both sides: ln(e²ˣ⁻¹) = ln(9)
Simplify left side: 2x − 1 = ln(9)
2x = 1 + ln(9)
x = (1 + ln(9))/2 ≈ (1 + 2.197)/2 ≈ 1.599
Use when: Result can be expressed as a logarithm to an unusual base
Steps
Example
Solve: 5ˣ = 20
Take log of both sides: log(5ˣ) = log(20)
Power rule: x · log(5) = log(20)
x = log(20)/log(5) = log₅(20)
x = ln(20)/ln(5) ≈ 2.996/1.609 ≈ 1.861
Use when: Both sides have exponential expressions with the same base
Steps
Example
Solve: 2ˣ⁺³ = 5ˣ
Take ln of both sides: (x + 3)ln(2) = x·ln(5)
Expand: x·ln(2) + 3·ln(2) = x·ln(5)
Group x terms: 3·ln(2) = x·ln(5) − x·ln(2)
Factor: 3·ln(2) = x(ln(5) − ln(2))
x = 3·ln(2)/(ln(5) − ln(2)) = 3ln(2)/ln(5/2) ≈ 2.079/0.916 ≈ 2.270
The change of base formula lets you evaluate any logarithm using ln or log₁₀, which every calculator has. It also gives the exact solution form for exponential equations with mismatched bases.
logb(x) = ln(x) / ln(b) = log(x) / log(b)
Valid for any b > 0, b ≠ 1, and x > 0
log₂(10)
ln(10)/ln(2)
≈ 3.322
log₃(7)
ln(7)/ln(3)
≈ 1.771
log₇(100)
ln(100)/ln(7)
≈ 2.366
Common Mistake
Wrong: log₃(7) = log(3)/log(7) — the fraction is inverted.
Right: log₃(7) = log(7)/log(3) — the argument goes on top, the base goes on the bottom.
These examples cover every major type of exponential equation. Work through each one before looking at the solution.
Matching Bases
Solve: 8ˣ = 2^(x+6)
Rewrite 8 as 2³: (2³)ˣ = 2^(x+6)
Simplify left: 2^(3x) = 2^(x+6)
Set exponents equal: 3x = x + 6
Subtract x: 2x = 6
x = 3
Answer: x = 3
Verify: 8³ = 512 = 2⁹ = 2^(3+6) ✓
Different Bases — Using ln
Solve: 2ˣ = 11, round to 4 decimal places
Take ln of both sides: ln(2ˣ) = ln(11)
Power rule: x · ln(2) = ln(11)
x = ln(11)/ln(2)
x = 2.3979/0.6931
x ≈ 3.4594
Answer: x ≈ 3.4594
Verify: 2^3.4594 ≈ 11 ✓
Natural Exponential
Solve: 3e^(2x) − 5 = 10
Isolate exponential: 3e^(2x) = 15
Divide by 3: e^(2x) = 5
Take ln: ln(e^(2x)) = ln(5)
Simplify: 2x = ln(5)
x = ln(5)/2 ≈ 0.8047
Answer: x = ln(5)/2 ≈ 0.8047
Verify: 3e^(2·0.8047) = 3e^1.609 ≈ 3 × 5 = 15 → 15 − 5 = 10 ✓
Quadratic Form — Substitution
Solve: 4ˣ − 3 · 2ˣ − 4 = 0
Note: 4ˣ = (2²)ˣ = (2ˣ)². Let u = 2ˣ
Substitute: u² − 3u − 4 = 0
Factor: (u − 4)(u + 1) = 0
u = 4 or u = −1
u = −1: 2ˣ = −1 impossible (discard — 2ˣ > 0 always)
u = 4: 2ˣ = 4 = 2² → x = 2
Answer: x = 2
Verify: 4² − 3·2² − 4 = 16 − 12 − 4 = 0 ✓
Compound Interest — Solve for Time
How long for $2,500 to grow to $5,000 at 6% annual rate, continuously compounded?
Formula: A = Pe^(rt)
Substitute: 5000 = 2500 · e^(0.06t)
Divide both sides by 2500: 2 = e^(0.06t)
Take ln: ln(2) = 0.06t
t = ln(2)/0.06 ≈ 0.6931/0.06
t ≈ 11.55 years
Answer: t ≈ 11.55 years
Rule of 72 estimate: 72/6 = 12 years — close to exact answer ✓
Pure exponential equations (bˣ = k) have no extraneous solutions when k > 0, because bˣ is always positive. However, the substitution technique (quadratic form) can produce candidates that must be screened:
Rule: Discard any solution where u = bˣ ≤ 0
Example: 9ˣ − 4 · 3ˣ − 5 = 0
Let u = 3ˣ → (3ˣ)² = 9ˣ, so u² − 4u − 5 = 0
Factor: (u − 5)(u + 1) = 0 → u = 5 or u = −1
u = −1: 3ˣ = −1 is impossible (3ˣ > 0 always) → DISCARD
u = 5: 3ˣ = 5 → x = ln(5)/ln(3) ≈ 1.465
Contrast with log equations
Logarithmic equations (log in the equation) frequently produce extraneous solutions because a log argument could become negative or zero. Always substitute answers back into log equations. Exponential equations are safer — but still check substitution results.
These three application types are tested heavily on precalculus exams and SAT/ACT. Each one reduces to solving an exponential equation using logarithms.
A = Pe^(rt) → t = ln(A/P) / r
$1,000 at 5%, continuously compounded. When does it reach $3,000?
3000 = 1000 · e^(0.05t)
3 = e^(0.05t)
ln(3) = 0.05t
t = ln(3)/0.05 ≈ 21.97 years
A(t) = A₀ · (1/2)^(t/h) → t = h · ln(fraction)/ln(0.5)
Carbon-14 half-life is 5,730 years. When does a sample reach 30% of its original amount?
0.30 = (1/2)^(t/5730)
ln(0.30) = (t/5730) · ln(0.5)
t = 5730 · ln(0.30)/ln(0.5)
t = 5730 · (−1.204)/(−0.693) ≈ 9,954 years
P(t) = P₀ · e^(kt) → doubling time = ln(2)/k
A bacteria colony grows at 8% per hour. Find the doubling time.
Set 2P₀ = P₀ · e^(0.08t)
Divide: 2 = e^(0.08t)
ln(2) = 0.08t
t = ln(2)/0.08 ≈ 8.66 hours
When both sides of the equation can be written with the same base, set the exponents equal and solve. For example, 2ˣ = 8 becomes 2ˣ = 2³, so x = 3. This works whenever the numbers are recognizable powers of a common base, such as 4, 8, 16, 27, 32, 125, etc. Rewrite both sides as powers of the same base, match exponents, and solve the resulting linear or polynomial equation.
Take the natural log (ln) or common log (log) of both sides, then use the power rule to bring the variable exponent down as a multiplier. For example, to solve 3ˣ = 7: take ln of both sides to get x·ln(3) = ln(7), then divide to get x = ln(7)/ln(3) ≈ 1.771. This is equivalent to the change of base formula: x = log₃(7) = ln(7)/ln(3).
The change of base formula converts any logarithm to natural log or common log: log_b(x) = ln(x)/ln(b) = log(x)/log(b). You use it when solving exponential equations with unusual bases (like 3ˣ = 7 → x = ln7/ln3) or when evaluating logarithms on a calculator that only has ln and log₁₀ buttons. The formula works for any base b > 0, b ≠ 1.
In compound interest A = Pe^(rt), solving for t when you know the target amount A gives an exponential equation: A/P = e^(rt). Take ln of both sides: ln(A/P) = rt, so t = ln(A/P)/r. For half-life A(t) = A₀·(1/2)^(t/h), set A(t)/A₀ equal to the desired fraction and solve for t using logarithms: t = h·ln(fraction)/ln(0.5). Population doubling time follows the same pattern: set 2P₀ = P₀·e^(kt) and solve to get t = ln(2)/k.
Pure exponential equations (where only the exponent contains the variable) do not produce extraneous solutions because bˣ > 0 for all real x and any base b > 0. However, when an exponential equation is rewritten in quadratic form using substitution (e.g., let u = 2ˣ), one solution for u may be negative or zero. Since bˣ > 0 always, any u ≤ 0 must be discarded. Always check substituted solutions by verifying that the recovered exponential base equation is satisfied.
Graphs, properties, base e, growth and decay
Log laws, solving log equations, change of base
Compound interest, half-life, Newton's cooling
Why ln is the inverse of eˣ and how inverses work
Complete study guide for all 13 chapters
50+ questions with step-by-step solutions
Interactive problems with step-by-step solutions and private tutoring. Chapter 4 covers all exponential and logarithmic equation types — free to try.
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