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Logarithmic Equations



Key Terms
Basic Form: Logarithm Equals a Constant
Logarithms on Both Sides
Combining Logarithms
Domain Restrictions
Extraneous Solutions
Equations Requiring Substitution
Solving Exponential Equations via Logarithms
Exponentials on Both Sides
Choosing Common or Natural Logarithm
Summary of Solving Methods
Logarithmic Equations FAQ



Isolating the Variable Inside the Logarithm

Equations containing logarithms require techniques that exploit the definition and properties of logarithmic functions. Converting to exponential form, applying the one-to-one property, and using logarithm rules to combine expressions — these methods reduce logarithmic equations to algebraic ones. Every solution must be verified against domain restrictions, as the solving process can introduce extraneous values.

Key Terms

Equation Concepts

Logarithmic Equation— an equation where the variable appears inside a logarithm
One-to-One Property— if log⁡a(M)=log⁡a(N)\log_a(M) = \log_a(N) then M=NM = N
Argument (of a Logarithm)— all arguments must be positive; solutions violating this are extraneous

Rules Used

Product Rule (Logarithms)— combining sums of logarithms
Power Rule (Logarithms)— moving coefficients into exponents
Change of Base Formula— solving exponential equations via logarithms

See All Algebra Definitions →


Basic Form: Logarithm Equals a Constant

The simplest logarithmic equation has the form log⁡a(x)=k\log_a(x) = k, where kk is a constant.

Convert to exponential form using the definition: log⁡a(x)=k\log_a(x) = k means ak=xa^k = x.

Example: Solve log⁡2(x)=5\log_2(x) = 5.
x=25=32x = 2^5 = 32


Example: Solve log⁡3(x)=−2\log_3(x) = -2.
x=3−2=19x = 3^{-2} = \frac{1}{9}


Example: Solve ln⁡(x)=4\ln(x) = 4.
x=e4≈54.598x = e^4 \approx 54.598


When the argument is an expression, solve the resulting equation:

Example: Solve log⁡5(2x−1)=2\log_5(2x - 1) = 2.
2x−1=52=252x - 1 = 5^2 = 25

2x=262x = 26

x=13x = 13


To solve your own logarithmic equation with each step shown, use the logarithmic equation solver.

Logarithms on Both Sides

When the same logarithm base appears on both sides, apply the one-to-one property: if log⁡a(M)=log⁡a(N)\log_a(M) = \log_a(N), then M=NM = N.

Example: Solve log⁡4(3x+2)=log⁡4(x+10)\log_4(3x + 2) = \log_4(x + 10).
3x+2=x+103x + 2 = x + 10

2x=82x = 8

x=4x = 4


Verification: log⁡4(3(4)+2)=log⁡4(14)\log_4(3(4) + 2) = \log_4(14) and log⁡4(4+10)=log⁡4(14)\log_4(4 + 10) = \log_4(14). Both arguments are positive, and the equation holds.

Example: Solve ln⁡(x2)=ln⁡(9)\ln(x^2) = \ln(9).
x2=9x^2 = 9

x=±3x = \pm 3


Check both: ln⁡((−3)2)=ln⁡(9)\ln((-3)^2) = \ln(9) works, and ln⁡((3)2)=ln⁡(9)\ln((3)^2) = \ln(9) works. Both solutions are valid.

Combining Logarithms

When multiple logarithms appear, use the rules to condense them into a single logarithm, then apply previous techniques.

Example: Solve log⁡2(x)+log⁡2(x−2)=3\log_2(x) + \log_2(x - 2) = 3.

Condense using the product rule:
log⁡2(x(x−2))=3\log_2(x(x - 2)) = 3

log⁡2(x2−2x)=3\log_2(x^2 - 2x) = 3


Convert to exponential form:
x2−2x=23=8x^2 - 2x = 2^3 = 8

x2−2x−8=0x^2 - 2x - 8 = 0

(x−4)(x+2)=0(x - 4)(x + 2) = 0

x=4 or x=−2x = 4 \text{ or } x = -2


Check domain: x=4x = 4 gives log⁡2(4)+log⁡2(2)\log_2(4) + \log_2(2), both defined. x=−2x = -2 gives log⁡2(−2)\log_2(-2), undefined. Reject x=−2x = -2.

Solution: x=4x = 4.
log2 x + log2(x - 2) = 3 only at x = 4; the curve exists for x > 2xy−3−2−1123456789−3−2−1123456log₂ x + log₂(x − 2)y = 3x = 4domain: x > 2x = −2 is rejected:log₂(−2) is undefinedOnly x = 4 survives: the left side exists only for x > 2
Why x = −2 is thrown out

The section's log₂ x + log₂(x − 2) = 3. The left side only exists for x > 2, where both arguments are positive, so its graph starts at the dashed line. It meets the level 3 once, at x = 4. Condensing to x² − 2x = 8 also produces x = −2, but no part of the graph lives there: log₂(−2) is undefined, so the candidate is rejected.

Checking each candidate against the domain is the step that cannot be skipped.

Domain Restrictions

Every argument of every logarithm in the original equation must be positive. Identify these restrictions before solving.

For log⁡a(f(x))\log_a(f(x)) to be defined, f(x)>0f(x) > 0.

Example: In log⁡3(x−5)+log⁡3(x+1)=2\log_3(x - 5) + \log_3(x + 1) = 2, the restrictions are:
x−5>0  ⟹  x>5x - 5 > 0 \implies x > 5

x+1>0  ⟹  x>−1x + 1 > 0 \implies x > -1


The intersection is x>5x > 5. Any solution must satisfy x>5x > 5.

Solving:
log⁡3((x−5)(x+1))=2\log_3((x-5)(x+1)) = 2

(x−5)(x+1)=9(x-5)(x+1) = 9

x2−4x−5=9x^2 - 4x - 5 = 9

x2−4x−14=0x^2 - 4x - 14 = 0

x=4±16+562=4±722=2±32x = \frac{4 \pm \sqrt{16 + 56}}{2} = \frac{4 \pm \sqrt{72}}{2} = 2 \pm 3\sqrt{2}


Since 32≈4.243\sqrt{2} \approx 4.24, we have x≈6.24x \approx 6.24 or x≈−2.24x \approx -2.24. Only x=2+32x = 2 + 3\sqrt{2} satisfies x>5x > 5.

Extraneous Solutions

Extraneous solutions arise when algebraic manipulation produces values that violate the original equation's domain. Always substitute back into the original equation.

Example: Solve log⁡(x)+log⁡(x−3)=1\log(x) + \log(x - 3) = 1.

log⁡(x(x−3))=1\log(x(x-3)) = 1

x(x−3)=10x(x-3) = 10

x2−3x−10=0x^2 - 3x - 10 = 0

(x−5)(x+2)=0(x-5)(x+2) = 0

x=5 or x=−2x = 5 \text{ or } x = -2


Check x=5x = 5: log⁡(5)+log⁡(2)\log(5) + \log(2) — both defined, sum equals log⁡(10)=1\log(10) = 1. Valid.

Check x=−2x = -2: log⁡(−2)\log(-2) — undefined. Extraneous.

The squaring, factoring, or combining steps do not "create" these solutions — they were always algebraic solutions to the transformed equation. The domain restrictions eliminate them.

The same discipline applies across every equation that combines logarithms. The table below collects four equations from this page that produce extra algebraic candidates and shows the domain check that eliminates the extraneous ones.
Equation Algebraic solutions Domain requirement Verdict after check
log₂(x) + log₂(x − 2) = 3 x = 4 or x = −2 x > 0 and x − 2 > 0 → x > 2 keep x = 4; reject x = −2 (extraneous)
log(x) + log(x − 3) = 1 x = 5 or x = −2 x > 0 and x − 3 > 0 → x > 3 keep x = 5; reject x = −2
log₃(x − 5) + log₃(x + 1) = 2 x = 2 ± 3√2 ≈ 6.24 or −2.24 x − 5 > 0 and x + 1 > 0 → x > 5 keep x = 2 + 3√2; reject the other
ln(x²) = ln(9) x = 3 or x = −3 x² > 0 → x ≠ 0 (both candidates fine) keep both x = 3 and x = −3

Equations Requiring Substitution

Some equations have a quadratic structure in the logarithm. Substitution simplifies the algebra.

Example: Solve (log⁡2(x))2−5log⁡2(x)+6=0(\log_2(x))^2 - 5\log_2(x) + 6 = 0.

Let u=log⁡2(x)u = \log_2(x):
u2−5u+6=0u^2 - 5u + 6 = 0

(u−2)(u−3)=0(u - 2)(u - 3) = 0

u=2 or u=3u = 2 \text{ or } u = 3


Back-substitute:
log⁡2(x)=2  ⟹  x=4\log_2(x) = 2 \implies x = 4

log⁡2(x)=3  ⟹  x=8\log_2(x) = 3 \implies x = 8


Both satisfy x>0x > 0. Solutions: x=4x = 4 and x=8x = 8.

Example: Solve ln⁡(x)+6ln⁡(x)=5\ln(x) + \frac{6}{\ln(x)} = 5.

Let u=ln⁡(x)u = \ln(x):
u+6u=5u + \frac{6}{u} = 5

u2+6=5uu^2 + 6 = 5u

u2−5u+6=0u^2 - 5u + 6 = 0

(u−2)(u−3)=0(u-2)(u-3) = 0


Back-substitute: x=e2x = e^2 or x=e3x = e^3.
(log2 x)^2 - 5 log2 x + 6 = 0 at x = 4 and x = 8xy12345678910−112345678(log₂ x)² − 5 log₂ x + 6u = log₂ x: u² − 5u + 6 = 0, so u = 2 or 3x = 2² = 4 or x = 2³ = 8Quadratic in log₂ x: zeros at x = 4 and x = 8
A quadratic hiding inside a logarithm

The section's (log₂ x)² − 5 log₂ x + 6 = 0. In terms of u = log₂ x it is u² − 5u + 6 = 0, with u = 2 or u = 3, so x = 4 or x = 8. The graph confirms both: it reaches zero exactly there, dipping slightly below the axis between them.

Each value of u gives one x, because log₂ x is one-to-one.

Solving Exponential Equations via Logarithms

When an exponential equation cannot be solved by matching bases, take the logarithm of both sides.

Example: Solve 3x=73^x = 7.

No integer power of 33 equals 77. Take logarithms:
log⁡(3x)=log⁡(7)\log(3^x) = \log(7)

x⋅log⁡(3)=log⁡(7)x \cdot \log(3) = \log(7)

x=log⁡(7)log⁡(3)≈0.8450.477≈1.771x = \frac{\log(7)}{\log(3)} \approx \frac{0.845}{0.477} \approx 1.771


Either log⁡\log or ln⁡\ln works — the ratio is identical.

Example: Solve 52x+1=125^{2x+1} = 12.

ln⁡(52x+1)=ln⁡(12)\ln(5^{2x+1}) = \ln(12)

(2x+1)ln⁡(5)=ln⁡(12)(2x+1)\ln(5) = \ln(12)

2x+1=ln⁡(12)ln⁡(5)2x + 1 = \frac{\ln(12)}{\ln(5)}

2x=ln⁡(12)ln⁡(5)−12x = \frac{\ln(12)}{\ln(5)} - 1

x=12(ln⁡(12)ln⁡(5)−1)≈0.271x = \frac{1}{2}\left(\frac{\ln(12)}{\ln(5)} - 1\right) \approx 0.271


To evaluate the final logarithm numerically, use the log calculator.

Exponentials on Both Sides

When exponential expressions with different bases appear on both sides, logarithms bring down both exponents.

Example: Solve 2x+3=5x−12^{x+3} = 5^{x-1}.

Take logarithms:
ln⁡(2x+3)=ln⁡(5x−1)\ln(2^{x+3}) = \ln(5^{x-1})

(x+3)ln⁡(2)=(x−1)ln⁡(5)(x+3)\ln(2) = (x-1)\ln(5)


Expand:
xln⁡(2)+3ln⁡(2)=xln⁡(5)−ln⁡(5)x\ln(2) + 3\ln(2) = x\ln(5) - \ln(5)


Collect xx terms:
xln⁡(2)−xln⁡(5)=−ln⁡(5)−3ln⁡(2)x\ln(2) - x\ln(5) = -\ln(5) - 3\ln(2)

x(ln⁡(2)−ln⁡(5))=−ln⁡(5)−3ln⁡(2)x(\ln(2) - \ln(5)) = -\ln(5) - 3\ln(2)

x=−ln⁡(5)−3ln⁡(2)ln⁡(2)−ln⁡(5)=ln⁡(5)+3ln⁡(2)ln⁡(5)−ln⁡(2)x = \frac{-\ln(5) - 3\ln(2)}{\ln(2) - \ln(5)} = \frac{\ln(5) + 3\ln(2)}{\ln(5) - \ln(2)}


Numerically: x=1.609+2.0791.609−0.693=3.6880.916≈4.026x = \frac{1.609 + 2.079}{1.609 - 0.693} = \frac{3.688}{0.916} \approx 4.026.
2^(x + 3) = 5^(x - 1) at x approximately 4.026xy12345501001502002503002ˣ⁺³5ˣ⁻¹x ≈ 4.026x = (ln 5 + 3 ln 2)/(ln 5 − ln 2) ≈ 4.0262ˣ⁺³ = 5ˣ⁻¹: the faster-growing 5ˣ⁻¹ overtakes at x ≈ 4.026
Two exponentials with different bases

The section's 2ˣ⁺³ = 5ˣ⁻¹. The left side starts higher, but base 5 grows faster, so the two curves cross exactly once. Taking logarithms of both sides brings the exponents down and gives x = (ln 5 + 3 ln 2)/(ln 5 − ln 2) ≈ 4.026, the crossing in the graph.

Either ln or log gives the same crossing; only the arithmetic differs.

Choosing Common or Natural Logarithm

    When taking logarithms to solve exponential equations, either base produces the same answer.

    For 3x=73^x = 7:
    x=log⁡(7)log⁡(3)=ln⁡(7)ln⁡(3)=log⁡3(7)x = \frac{\log(7)}{\log(3)} = \frac{\ln(7)}{\ln(3)} = \log_3(7)


    All three expressions are equal by the change of base formula.

    Practical considerations:

  • If the base is ee, use ln⁡\ln: solving e2x=5e^{2x} = 5 gives 2x=ln⁡(5)2x = \ln(5), no division needed.

  • If the base is 1010, use log⁡\log: solving 10x−1=5010^{x-1} = 50 gives x−1=log⁡(50)x - 1 = \log(50).

  • Otherwise, either works. Choose based on calculator availability or notational preference.

    • The expressions log⁡(7)/log⁡(3)\log(7)/\log(3) and ln⁡(7)/ln⁡(3)\ln(7)/\ln(3) compute to identical decimal values.
Exponential form Recommended logarithm Resulting expression for x
aˣ = B (any positive base a) either log or ln x = log(B) ⁄ log(a) = ln(B) ⁄ ln(a)
eˣ = B (base e) ln (cleaner — no division) x = ln(B)
10ˣ = B (base 10) log (cleaner — no division) x = log(B)
aᵐˣ⁺ⁿ = bᵖˣ⁺ʳ (different bases) either; ln conventional expand (mx+n)·ln(a) = (px+r)·ln(b), then solve a linear equation in x

Summary of Solving Methods

The page covers two related families of equations: equations whose unknown sits inside a logarithm, and exponential equations solved by taking a logarithm of both sides. The same toolkit — convert, condense, apply the one-to-one property, substitute, and check the domain — handles every form. The table below collects each equation form with its method and a worked example, in the order the page introduces them.
Equation form Method Worked example
logₐ(x) = k convert to exponential form: x = aᵏ log₂(x) = 5 → x = 32
logₐ(M) = logₐ(N) apply the one-to-one property: M = N log₄(3x+2) = log₄(x+10) → x = 4
Sum or difference of logs = k condense with logarithm rules, then convert log₂(x) + log₂(x−2) = 3 → x = 4
Quadratic in logₐ(x) substitute u = logₐ(x), solve, back-substitute (log₂x)² − 5log₂x + 6 = 0 → x = 4 or 8
aˣ = B (no matching bases) take log or ln of both sides, then divide 3ˣ = 7 → x = log(7) ⁄ log(3) ≈ 1.771
aᵐˣ⁺ⁿ = bᵖˣ⁺ʳ take logs, expand, collect x terms, solve linear equation 2ˣ⁺³ = 5ˣ⁻¹ → x ≈ 4.026

Logarithmic Equations FAQ

Do you always reject negative solutions in logarithmic equations?

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No. What has to be positive is each argument, not the value of xx itself. Solving ln⁡(x2)=ln⁡(9)\ln(x^2) = \ln(9) gives x=±3x = \pm 3, and both are valid because (−3)2=9(-3)^2 = 9 is positive. Contrast log⁡(x)+log⁡(x−3)=1\log(x) + \log(x - 3) = 1, where x=−2x = -2 fails only because it makes log⁡(−2)\log(-2) undefined.Read more →

Where do extraneous solutions in logarithmic equations come from?

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Condensing widens the domain. In log⁡3(x−5)+log⁡3(x+1)=2\log_3(x - 5) + \log_3(x + 1) = 2 the original requires x>5x > 5, but the condensed form log⁡3((x−5)(x+1))=2\log_3((x-5)(x+1)) = 2 only requires the product to be positive, which also permits x<−1x < -1. The algebra invents nothing; the extra root genuinely solves the condensed equation, and the original domain is what rules it out.Read more →

Do substitution problems like (log⁡2x)2−5log⁡2x+6=0(\log_2 x)^2 - 5\log_2 x + 6 = 0 need a domain check?

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Rarely, because back-substitution takes care of it. Solving log⁡2x=u\log_2 x = u gives x=2ux = 2^u, which is positive for every real uu, so the domain condition holds automatically and nothing is rejected. Substitution does not widen the domain the way condensing does. The exception is a substitution that divides, as in ln⁡x+6ln⁡x=5\ln x + \frac{6}{\ln x} = 5, which needs ln⁡x≠0\ln x \neq 0.Read more →