Natural Log Solver

Equation Setup

ln(x) = 3

a (coeff)

b (const)

c (RHS)

Solve ln(ax + b) = c → ax + b = e^c → x = (e^c − b) / a

Examples

e^x = 20

a (coeff)

b (const)

c (RHS)

Solve e^ax+b = c → ax + b = ln(c) → x = (ln(c) − b) / a

Examples

2·ln(x) = ln(x) + 3

a (left coeff)

b (right coeff)

c (const)

Solve a·ln(x) = b·ln(x) + c. Collect ln terms: (a−b)·ln(x) = c → x = e^(c/(a−b))

Examples

Solution

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

Solution

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Exact Form

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Domain Check

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Verification

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Inverse Operations

e^(ln x) = x and ln(e^x) = x

ln and e are inverses of each other. To undo a natural log, raise e to both sides. To undo e raised to a power, take ln of both sides.

Key values to memorize:

ln(1) = 0 (because e⁰ = 1)
ln(e) = 1 (because e¹ = e)
ln(e²) = 2 (because e² = e²)
ln(eⁿ) = n for any n

When solving ln(ax+b) = c: raise e to both sides to get ax+b = e^c, then solve the linear equation.

Always check your answer satisfies the domain: the argument of ln must be strictly positive.

Domain Restriction

ln(u) is defined only when u > 0

The natural log ln(x) is only defined for x > 0. You cannot take the log of zero or a negative number.

When solving ln(ax+b) = c, your final answer x must satisfy ax+b > 0. If it doesn't, the equation has no real solution.

For e^(ax+b) = c: since eˣ is always positive, c must be positive for any real solution to exist (ln of a negative is undefined).