About the Exponential Equation Solver
This exponential equation solver finds x when the unknown is in the exponent. Choose the form a·b^(kx + m) + d = c — which covers everything from 2ˣ = 32 and 3·2ˣ = 48 to e^(2x − 1) = 10 and 500·1.06ˣ = 1000 — or a·bˣ = c·qˣ, where both sides are exponentials with different bases. The calculator isolates the power, takes logarithms of both sides and returns x as a decimal and as an exact logarithm expression.
It is built for algebra 2 and pre-calculus students learning to solve exponential equations with logs, and for practical questions such as how many years an investment takes to double, when a population reaches a target, or how long until a radioactive sample decays to a given level.
Every step is written out (subtract, divide, take the log, solve the linear equation) and the answer is substituted back to check it. If the equation has no real solution — for example when the isolated power would have to be negative — the calculator explains why. For base e type 2.718281828.
With the default inputs, the solution x is 4. Change any value above to recalculate instantly.
How to use the exponential equation solver
- 1Choose the form that matches your equation.
- 2Enter the coefficient a and base b (2.718281828 for e).
- 3For a·b^(kx + m) + d = c, enter k, m, d and c; for two bases, enter c and q.
- 4Read x, the exact logarithm form and the steps.
- 5Use the check value and graph to confirm the sides are equal at x.
Formula and method
To solve an exponential equation, first isolate the power: subtract d and divide by a so that b^(kx + m) = (c − d)/a. Because the logarithm undoes exponentiation, take the log of both sides: kx + m = log_b((c − d)/a), and compute log_b with the change-of-base rule ln(y) ÷ ln(b). Then solve the linear equation for x. A real solution exists only when (c − d)/a > 0, since a positive base to any real power is positive.
With exponentials on both sides, a·bˣ = c·qˣ, take natural logs: ln a + x ln b = ln c + x ln q. Collect the x terms to get x = ln(c/a) ÷ ln(b/q). If the bases are equal the x terms cancel and there is either no solution or infinitely many.
- a
- Coefficient in front of the power
- b, q
- Bases (positive, not 1)
- k, m
- Exponent is kx + m
- c, d
- Right-hand side and added constant
Worked examples
3 · 2ˣ = 48
Divide by 3 to get 2ˣ = 16. Taking logs gives x = ln 16 ÷ ln 2 = 4, which you can confirm because 2⁴ = 16 and 3 × 16 = 48.
Doubling time: 5000 · 1.06ˣ = 10000
An investment growing 6% a year doubles when 1.06ˣ = 2, so x = ln 2 ÷ ln 1.06 ≈ 11.9 years — close to the rule-of-72 estimate of 72 ÷ 6 = 12.
Base e: e^(2x − 1) = 10
Taking the natural log gives 2x − 1 = ln 10 ≈ 2.302585, so 2x = 3.302585 and x ≈ 1.6513.
Two bases: 4 · 2ˣ = 3ˣ
Logs give ln 4 + x ln 2 = x ln 3, so x = ln 4 ÷ (ln 3 − ln 2) = ln(1/4) ÷ ln(2/3) ≈ 3.419. At that x both sides equal about 42.8.
Frequently asked questions
How do you solve an exponential equation?+
Isolate the exponential part (b^(…) = number), take the logarithm of both sides, and solve the resulting linear equation. For example 2ˣ = 20 gives x = ln 20 ÷ ln 2 ≈ 4.32.
When can I solve without logarithms?+
When both sides can be written as powers of the same base. 9ˣ = 27 becomes 3²ˣ = 3³, so 2x = 3 and x = 1.5. Otherwise logarithms are needed.
Does it matter whether I use ln or log?+
No. ln y ÷ ln b and log₁₀ y ÷ log₁₀ b give the same value of log_b y (the change-of-base rule). Natural logs are used here because they appear in growth and decay formulas.
Why does an exponential equation have no solution?+
A positive base raised to any real power is always positive. If isolating the power gives bˣ = 0 or a negative number — for example 2ˣ + 5 = 3 — there is no real x that works.
How do I find a doubling or half-life time?+
Doubling time solves (1 + r)ᵗ = 2, so t = ln 2 ÷ ln(1 + r). A half-life solves bᵗ = 0.5. Enter a = 1, the growth factor as b and c = 2 (or 0.5) in this solver.