Exponential & Logarithmic Equation Solver
Solve all standard exponential and logarithmic equations — a·bˣ=c, eˣ=c, log₂(x)=c, ln(ax+b)=c, and quadratic-in-exponential forms — with complete step-by-step algebraic working.
What Is the Exponential & Logarithmic Equation Solver?
Exponential equations have the variable in the exponent (bˣ = c), while logarithmic equations have the variable inside a logarithm (log x = c). Both are solved by applying the inverse operation: take a logarithm of both sides to “bring down” an exponent, or exponentiate both sides to “undo” a logarithm.
The change-of-base formula (log_b x = ln x / ln b) converts any base to natural logarithms, which is what calculators and computers compute natively. For the quadratic-in-exponential form, the substitution u = aˣ converts the equation into a standard quadratic in u — only positive u values give real x solutions, since aˣ > 0 for any real x when a > 0.
Formula
Core rule: bˣ = c → x = log_b(c) = ln(c)/ln(b)
Equation forms:
a·bˣ = c → bˣ = c/a → x = ln(c/a)/ln(b)
a·e^(bx+d) = c → bx+d = ln(c/a) → x = (ln(c/a)−d)/b
log_n(x) = c → x = nᶜ
ln(ax+b) = c → ax+b = eᶜ → x = (eᶜ−b)/a
log_n(ax+b) = c → ax+b = nᶜ → x = (nᶜ−b)/a
Quadratic in exponential:
A·(aˣ)² + B·aˣ + C = 0 → let u = aˣ, solve Au²+Bu+C=0, then x = ln(u)/ln(a)
How to Use
- 1Choose equation type: Select from the six forms — pick the one that matches your equation's structure.
- 2Enter parameters: Fill in a, b, c, and any other constants shown for that equation type.
- 3Solve: Click "Solve". The solution appears with the exact numeric value.
- 4Read steps: The numbered step-by-step panel shows every algebraic operation: dividing, taking logs, solving for x.
- 5Check no-solution cases: If the equation has no real solution (e.g., bˣ = negative), the calculator explains why rather than returning an error.
Example Calculation
Example 1 — a·bˣ = c: 2·3ˣ = 18
- ›Divide by 2: 3ˣ = 9
- ›Take log: x·ln(3) = ln(9)
- ›x = ln(9)/ln(3) = 2.197/1.099 = 2
- ›Verify: 2·3² = 2·9 = 18 ✓
Example 2 — Quadratic: (2ˣ)² − 5·2ˣ + 6 = 0
- ›Let u = 2ˣ: u² − 5u + 6 = 0
- ›Discriminant = 25 − 24 = 1 → u = (5±1)/2 → u₁=3, u₂=2
- ›u₁=3 > 0: x₁ = ln(3)/ln(2) ≈ 1.585
- ›u₂=2 > 0: x₂ = ln(2)/ln(2) = 1
- ›Two real solutions: x ≈ 1.585 and x = 1
Understanding Exponential & Logarithmic Equation
Equation Types at a Glance
| Form | Key step | Solution formula |
|---|---|---|
| a·bˣ = c | Divide by a, take log_b | x = ln(c/a) / ln(b) |
| a·e^(bx+d) = c | Divide by a, take ln | x = (ln(c/a) − d) / b |
| log_n(x) = c | Exponentiate base n | x = nᶜ |
| ln(ax+b) = c | Exponentiate base e | x = (eᶜ − b) / a |
| log_n(ax+b) = c | Exponentiate base n | x = (nᶜ − b) / a |
| A·(aˣ)²+B·aˣ+C=0 | Substitute u=aˣ, quadratic | x = ln(u)/ln(a) for u > 0 |
Applications
Exponential and logarithmic equations arise in compound interest (A = P·eʳᵗ, solve for t), radioactive decay (N = N₀·e^(−λt), solve for t), pH calculations (pH = −log₁₀[H⁺], solve for [H⁺]), decibels (dB = 10·log₁₀(P/P₀), solve for P), and earthquake magnitude (Richter scale). The “doubling time” formula t = ln(2)/r is a direct application of solving 2 = e^(rt).
Frequently Asked Questions
Why is bˣ always positive?
For b > 0, bˣ = e^(x·ln b) > 0 always, since eˣ > 0 for all real x. An equation like bˣ = negative number has no real solution. The calculator detects and explains this automatically rather than returning an error.
What is the change-of-base formula and when is it needed?
log_b(x) = ln(x)/ln(b). This converts any base to natural log, which computers calculate directly. For example, log₅(100) = ln(100)/ln(5) ≈ 2.861. The solver applies this automatically for all base-b equation types.
How do I solve an equation where the variable appears in two different exponents?
Convert both sides to the same base if possible: 4ˣ = 8^(x−1) → 2^(2x) = 2^(3x−3) → 2x = 3x−3 → x=3. If no common base exists, take ln of both sides: x·ln(a) = f(x)·ln(b), then solve the resulting linear or algebraic equation.
Why must the argument of a logarithm be positive?
The function ln(x) is defined only for x > 0. After solving ln(ax+b) = c for x, substitute back to verify ax+b > 0. The calculator performs this check and flags any solution where the logarithm argument is non-positive.
You Might Also Like
Explore 360+ Free Calculators
From math and science to finance and everyday life — all free, no account needed.