Logarithmic Equation Solver

Equation Setup

log(x − 2) = 1

base b

a (coeff)

c (const)

d (RHS)

Solve log₂(ax + c) = d → ax + c = b^d → x = (b^d − c) / a. Domain: ax + c > 0.

Examples

log(2x + 1) = log(x + 5)

base b

a₁ (left coeff)

c₁ (left const)

a₂ (right coeff)

c₂ (right const)

Since bases match: f(x) = g(x). Set a₁x + c₁ = a₂x + c₂ and solve.

Examples

log(x) + log(x − 3) = 1

base b

d (RHS)

c₁ (first log const)

c₂ (second log const)

Condenses log(x + c₁) + log(x + c₂) = d → log((x+c₁)(x+c₂)) = d → solve the quadratic.

Examples

Solution

Enter values above and press Solve to see the solution, exact form, domain check, and verification.

Solution

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Exact Form

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Domain Check

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Verification

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Steps for Log Equations

log₂(u) = d ⟺ u = b^d

1. Isolate the logarithm on one side of the equation.

2. Convert to exponential form. If log₂(u) = d, then u = b^d. This removes the log completely.

3. Solve the resulting equation (linear or quadratic) for x.

4. Check the domain. Substitute x back — the argument of every log must be strictly positive.

Always convert to exponential form to undo a logarithm — it is the single most important log-equation strategy.

Domain Restriction

log₂(u) defined only when u > 0

The argument of any logarithm must be strictly positive. You cannot take the log of zero or a negative number.

When solving log equations — especially ones that produce a quadratic — always substitute every candidate solution back into the original equation to verify the argument is positive. Extraneous roots appear when squaring or multiplying through.

Check ax + c > 0 after solving Tab 1.
Check both a₁x + c₁ > 0 and a₂x + c₂ > 0 for Tab 2.
For Tab 3: check both x + c₁ > 0 and x + c₂ > 0. Reject any root that fails.
A negative root of the quadratic is often extraneous — confirm before accepting.

Log equations still tricky?