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Exponential and logarithmic equations

Solve exponential and log equations with common bases, logarithms, quadratic substitutions and domain checks. Includes natural logs, reciprocal exponentials and mixed bases.

Before you startIndex and log laws, quadratics and the inverse relationship between ln and exp

01 / Choose a method and record the domain

Algebraic candidates are not automatically valid solutions.

Start by checking that every logarithm input is positive. Then decide whether to use a common base, take logarithms, combine logarithms, or substitute for a repeated exponential or logarithm.

Taking logs of both sides is valid when both sides are positive. Combining logs also keeps the conditions of each original input; a positive product alone does not guarantee that both factors were positive.

After solving, return to the original equation and check its domain. State all valid solutions, using exact forms before rounded decimals.

02 / Use a common base when possible

Equal powers of the same valid base have equal exponents.

For a > 0 and a ≠ 1, aᵘ = aᵛ implies u = v because the exponential is one-to-one.

4^x+1 = 8^x−1
2^2x+2 = 2^3x−3
2x + 2 = 3x − 3 ⇒ x = 5

An identity or a contradiction

2^2x = 4ˣ holds for every real x. But 2^2x = 2 · 4ˣ has no solution, because 4ˣ is positive and division gives 1 = 2. Cancelling a variable does not always leave one numerical answer.

03 / Take logs to bring down an exponent

Use brackets around the whole exponent.

5^2x−1 = 18
(2x − 1)ln 5 = ln 18
x = ½[1 + ln(18)/ln(5)] ≈ 1.39794

Any consistent valid log base gives the same result. Taking ln is often convenient; the base of the exponential need not be e.

A base below one

(0.8)ˣ = 0.2
x = ln(0.2)/ln(0.8) ≈ 7.21257

Both logarithms are negative, so the ratio is positive. This is consistent with repeated decay reaching a smaller positive quantity.

04 / Different bases on both sides

Collect every term containing x after taking logs.

3^x+2 = 5^2x−1
(x + 2)ln 3 = (2x − 1)ln 5
x(2ln 5 − ln 3) = ln 5 + 2ln 3
x = (ln 5 + 2ln 3)/(2ln 5 − ln 3)
≈ 1.79537

Do not equate the exponents when their bases differ. The logarithm factors are what connect the two scales.

05 / A quadratic in an exponential

Substitute once, solve, then undo the substitution.

For e^2x − 5eˣ + 4 = 0, use u = eˣ, so u > 0:

u² − 5u + 4 = 0
(u − 1)(u − 4) = 0
u = 1 or 4
x = 0 or ln 4

Both values of u are positive, so both give real x. The model compares this with equations where a quadratic root must be rejected. It shows all candidates together so you can inspect the reason for each decision.

Keep the valid solutions in viewMove at your pace

Equation

e²ˣ − 5eˣ + 4 = 0. u = eˣ > 0. Candidates: u = 1 — valid, x = 0; u = 4 — valid, x = ln 4. Both substituted values are positive.

Watch a quadratic substitution retain only positive exponential values

Pause, replay or seek freely. The notes explain the same idea and stay in view.

06 / Rewrite shifted exponents first

A constant in the exponent becomes a multiplier.

2^2x+1 − 9 · 2ˣ + 4 = 0
Let u = 2ˣ > 0:
2u² − 9u + 4 = 0
(2u − 1)(u − 4) = 0
u = 1/2 or 4 ⇒ x = −1 or 2

The first term is 2(2ˣ)², not (2ˣ)² + 1. Index laws must be applied before treating the expression as a quadratic.

07 / A reciprocal exponential

Clear the denominator using a factor that is always positive.

eˣ + 6e⁻ˣ = 5
Multiply by eˣ:
e^2x − 5eˣ + 6 = 0
(eˣ − 2)(eˣ − 3) = 0
x = ln 2 or ln 3

Multiplying by eˣ loses no solutions and introduces no zero-factor case because eˣ is never zero.

08 / A sum of logarithms

Keep the original positive-input conditions.

Solve ln(x − 1) + ln(x + 1) = ln 8. The original domain is x > 1.

ln[(x − 1)(x + 1)] = ln 8
x² − 1 = 8
x = ±3

Only x = 3 is allowed. At x = −3 both original log inputs are negative, even though their product is positive. The simplified product equation alone has a larger domain.

09 / A difference of logarithms

Divide inputs before undoing the logarithm.

For log₃(x + 6) − log₃(x − 2) = 2, the domain is x > 2:

log₃[(x + 6)/(x − 2)] = 2
x + 6 = 9(x − 2)
x = 3

This satisfies the domain and gives log₃(9) = 2 after combining. Replacing the left side by log₃[(x + 6) − (x − 2)] would be an invalid subtraction rule.

10 / A quadratic in ln(x)

The substituted logarithm may be negative.

(ln x)² − ln x − 6 = 0, x > 0
Let u = ln x:
(u − 3)(u + 2) = 0
u = 3 or −2
x = e³ or e⁻²

Unlike u = eˣ, the substitution u = ln(x) allows every real u. Both resulting x-values are positive, so both are valid.

11 / Convert different logarithm bases

Do not cancel logarithms with unequal bases.

Solve log₂(x + 1) = log₄(3x + 1). Both inputs must be positive, giving x > −1/3. Since log₄(N) = ½log₂(N):

2log₂(x + 1) = log₂(3x + 1)
(x + 1)² = 3x + 1
x(x − 1) = 0
x = 0 or 1

Both pass the original domain check. Directly equating x + 1 with 3x + 1 would have lost the second solution.

12 / Diagnose an incorrect log step

A quotient, not a difference, comes out of subtraction.

Consider log₂(x) − ½log₂(x + 3) = 1. The domain is x > 0. Correct combination gives:

log₂[x/√(x + 3)] = 1
x/√(x + 3) = 2
x² = 4(x + 3)
(x − 6)(x + 2) = 0

Only x = 6 is valid. The incorrect step x − √(x + 3) = 2 confuses the quotient law with subtraction of inputs. Squaring also requires returning to the original sign and domain conditions.

13 / A product of different exponentials

Taking ln turns both factors into linear terms.

3ˣe^2x−1 = 7
x ln 3 + 2x − 1 = ln 7
x = (1 + ln 7)/(2 + ln 3) ≈ 0.950719

Both factors are positive for every real x, so taking ln is valid. Keep the whole exponent 2x − 1 intact when applying ln(e^2x−1).

a,b > 0, a > b, a + b = 10 and log₃(a) + log₃(b) = 2. Find a,b.

Hint

Turn the logarithm equation into ab = 9.

Worked solution

a = 9, b = 1.

15 / Recap

Check the equation you started with.

Use a common base or take logs of positive expressions.
Expand shifted exponents before making a substitution.
An exponential substitution is positive; a logarithm substitution may be any real number.
Retain each original log-input condition after combining.
Convert unequal log bases before equating their inputs.
Check every candidate, including any introduced by squaring.

Next: logarithmic graphs for modelling →

Section 1 of 15 · Choose a method and record the domain