Exponential And Logarithmic Equations Calculator
An unknown sitting in an exponent or inside a logarithm needs a different move from ordinary algebra: you undo the exponential with a logarithm, or undo the logarithm by exponentiating. This solver handles the six standard forms: solve ax = b, solve loga(x) = b, solve ex = b, solve ln(x) = b, and evaluate loga(b) or ax directly. Every answer names the logarithm law used at each step, flags exact integer answers when they exist, and checks itself by substituting the result back into the original equation.
Solves a^x = b
Domain rules enforced: bases must be positive and not 1; results of a^x and arguments of logs must be positive.
Quick Answer
For the pre-filled equation 3x = 81, the solver returns x = 4 exactly (81 = 34), with the check 34.000000 = 81.000000. When no exact power exists it divides logarithms instead: 2x = 10 gives x = ln(10) / ln(2) = 3.321928 at the default 6 decimal places. Pick any of the six modes, enter the two numbers, and the steps, the answer and the substitution check all update together.
Key takeaways
- Exponential equations fall to logarithms: ax = b becomes x = ln(b) / ln(a). Logarithmic equations fall to exponentials: loga(x) = b becomes x = ab.
- Exact answers come first. If b is an exact integer power of a, the solver says so instead of burying the integer in decimals.
- No solution is an answer too. If b is 0 or negative in ax = b, the solver stops with the domain reason rather than printing a meaningless number.
- The check line is part of the answer: every result is substituted back into the original equation before you trust it.
- Quick Answer
- How to use the equations solver
- The three laws behind every mode
- Worked example: solving 3^x = 81 step by step
- Six modes, six real runs
- What this solver does not do, and how to bridge the gap
- Same-base method or logarithms: which one first?
- What competing calculators miss
- Frequently asked questions
- Add This Calculator to Your Website
How to use the equations solver
Choose the form, enter the two numbers it asks for, then click Solve. The labels rename themselves for each mode, so you always know which number is which.
- Pick the equation type: the four Solve modes find x; the two Evaluate modes work out loga(b) or ax directly, which is what you want when there is no unknown to find.
- Enter base a and the second value: in Solve ax = b, a is the base and b is the result. In the ex and ln(x) modes the base is fixed at e, so the base box hides and only b is needed.
- Set the decimal places: 2, 4, 6 or 8. This controls the answer, the steps and the check together. Exact answers ignore it.
- Read all three lines: the answer, the steps naming the law used, and the Check line that substitutes the answer back. If the check does not reproduce your b, something was entered in the wrong box.
| Mode | You enter | It returns | Move it uses |
|---|---|---|---|
| Solve a^x = b | Base a, result b | x | Log of both sides, power rule |
| Solve log_a(x) = b | Base a, result b | x | Rewrite as x = a^b |
| Evaluate log_a(b) | Base a, value b | The logarithm | Change of base: ln(b) / ln(a) |
| Evaluate a^x | Base a, exponent x | The power | Direct evaluation |
| Solve e^x = b | Result b only | x = ln(b) | Natural log of both sides |
| Solve ln(x) = b | Result b only | x = e^b | Rewrite as x = e^b |
Solving and evaluating are inverse jobs, and students mix them up constantly. If your problem already gives you a, b and x and asks for the value of the expression, you are evaluating, not solving. If a function value is what you need for a range of inputs, the exponential function formula calculator evaluates y = a · bx and y = a · ekx directly.
The three laws behind every mode
Three logarithm laws do all the work in this solver, and the steps line names the one being used so you can follow the method, not just the answer.
Change of base: loga(b) = ln(b) / ln(a), which evaluates a log in any base using the natural log.
Inverse relationship: loga(x) = b exactly when ab = x, which converts logarithmic equations to exponential ones and back.
The power rule is why Solve ax = b works at all: take ln of both sides, the x drops out of the exponent, and dividing by ln(a) isolates it. The inverse relationship is the same fact read in the other direction, and it is the whole method for the log modes: log5(x) = 3 and 53 = x are one statement in two notations. If converting between those two notations is the part that trips you, the convert to exponential form calculator drills exactly that single step. The natural modes are not separate mathematics: ln is log base e, so ex = b is ax = b with a fixed at e, and the solver gives it its own mode only because it shows up so often in growth and decay work.
Worked example: solving 3^x = 81 step by step
The pre-filled example solves 3x = 81, and it shows the solver’s two-layer answer: an exact result when one exists, and the general logarithm method underneath it.
The steps the solver prints are the method you would use when there is no exact power: take the natural log of both sides, ln(3x) = ln(81); apply the power rule, x · ln(3) = ln(81); then x = ln(81) / ln(3) = 4.39444915 / 1.09861229, which is 4. Run the same mode with a = 2 and b = 10 and the exact layer stays silent, because 10 is not a power of 2: the answer is x = 3.321928, and the check line confirms 23.321928 = 10.000000. Both runs are real outputs of the tool above at 6 decimal places.
Six modes, six real runs
Each row below is an actual run of this solver at the default precision, including the two domain errors students hit most. Type any row in and you will get the same line back.
| Run | Answer line | Check line |
|---|---|---|
| Solve log_5(x) = 3 | x = 125.000000 | log_5(125.000000) = 3.000000 |
| Evaluate log_2(32) | log_2(32) = 5 exactly | 2^5.000000 = 32.000000 |
| Evaluate 9^0.5 | 9^0.5 = 3.000000 | No check needed: direct power |
| Solve e^x = 20 | x = 2.995732 | e^2.995732 = 20.000000 |
| Solve ln(x) = 2 | x = 7.389056 | ln(7.389056) = 2.000000 |
| Solve 2^x = 0 or 2^x = -8 | Error: “For a^x = b, b must be greater than zero.” A positive base raised to a real power is always positive, so these equations have no real solution. | |
Two edge behaviours are worth knowing before an exam does it for you. A fractional base works fine: 0.5x = 0.25 gives x = 2.000000, and 4x = 2 gives x = 0.500000, because the exact-power shortcut only fires for integer bases and the logarithm method handles everything else, as any exponent calculator run will confirm. And a negative result for the log modes is legitimate: log2(x) = -3 gives x = 0.125000, a positive argument with a negative logarithm, which is exactly what bases above 1 do to numbers between 0 and 1.
What this solver does not do, and how to bridge the gap
This solver handles the six clean forms above, and it is honest about the boundary: three common equation shapes sit outside it, because each needs an algebra step before one of the six modes applies.
- Something added inside the log or the exponent, such as log3(x + 1) = 2 or 52x – 1 = 17. Bridge: undo the shift by hand first. For log3(x + 1) = 2, the mode gives the inside value, 32 = 9, then x = 9 – 1 = 8. For 52x – 1 = 17, solve 5y = 17 in the solver (y = 1.760374), then 2x – 1 = 1.760374, so x = 1.380187.
- x on both sides, such as 5x + 2 = 4x. Bridge: take ln of both sides and collect the x terms: (x + 2) ln(5) = x ln(4), so x = -2 ln(5) / ln(5/4), about -14.425135. No single mode here does that collection step for you.
- Sums or differences of logarithms, such as ln(x) + ln(x – 1) = ln(6). Bridge: combine with the product rule first, ln(x(x – 1)) = ln(6), so x(x – 1) = 6, a quadratic: x = 3 (the x = -2 root is rejected because it makes ln(x) undefined, an extraneous solution).
That last point is the one that costs the most marks. Any time you exponentiate or combine logarithms, substitute the candidate answers back and reject any that make a logarithm’s argument zero or negative. The solver’s Check line is that habit built into the tool: for the forms it handles, the argument is x itself and the check is automatic.
Same-base method or logarithms: which one first?
Textbooks teach two routes into an exponential equation, and picking the right one first saves time. The solver always uses logarithms, but by hand the same-base route is faster when it works.
Same-base method
Rewrite both sides as powers of one base and set the exponents equal: 4x – 1 = 82x becomes 22x – 2 = 26x, so 2x – 2 = 6x and x = -0.5.
- Fastest when b is a clean power of a
- Gives exact answers with no decimals
- Fails completely when no common base exists
Logarithm method (this solver)
Take ln of both sides and use the power rule: 2x = 10 becomes x = ln(10) / ln(2) = 3.321928. Works for every valid a and b.
- Works whether or not a common base exists
- The solver still flags the exact answer when one exists
- Decimal answers need a sensible precision choice
The practical rule: glance at b first. If it is 4, 8, 9, 25, 27, 32, 81, 125, 1000 and friends, the same-base route is one line. Anything else, go straight to logarithms. In growth and decay problems the unknown is usually time, which is why this pattern matters beyond the classroom: solving ax = b for x is exactly the step the exponential growth calculator runs in reverse when you ask how long a quantity takes to reach a target.
What competing calculators miss
The solver landscape splits into tools that accept anything and teach little, and guides that teach everything and compute nothing. The gaps are consistent:
- Free-text solvers (Symbolab, SageCalculator): you type an equation as text and they parse it, which handles shifted arguments and x on both sides, but a single typo silently changes the problem, and the full step-by-step working sits behind an account or a paywall on the biggest of them. Structured modes cannot be mistyped.
- Single-form tools (CalculatorSoup Solve for Exponents, its Logarithm Calculator): each handles one shape, xn = y or one log relationship, cleanly and with good explanations. Six separate bookmarks replace what is one solver here.
- Textbook guides (OpenStax and LibreTexts, Albert): the method teaching is excellent, including extraneous-solution checks, but the arithmetic is left to you, and there is no exact-answer detection or substitution check on your numbers.
- The gap this page fills: all six standard forms in one tool, the law named at each step, exact integer answers surfaced instead of decimalised, a substitution check on every answer, and an honest list of the three shapes it does not handle with the manual bridge for each.
Frequently asked questions
Take the natural log of both sides and use the power rule to bring the exponent down: x times ln(3) = ln(81), so x = ln(81) / ln(3) = 4. Because 81 is an exact power of 3, this solver reports x = 4 exactly and checks it with 3^4.000000 = 81.000000.
Rewrite it in exponential form using the inverse relationship: log base 5 of x equals 3 exactly when x = 5^3. The solver returns x = 125.000000 and checks it by confirming log_5(125.000000) = 3.000000. For ln(x) = b the same move gives x = e^b.
Three shapes sit outside its six modes: something added inside the log or the exponent, such as log_3(x + 1) = 2; x on both sides, such as 5^(x + 2) = 4^x; and sums of logarithms, such as ln(x) + ln(x – 1) = ln(6). Each needs one algebra step by hand first, and the scope section below shows the bridge for all three.
Before computing decimals, it checks whether b is an exact integer power of a: 81 = 3^4, 32 = 2^5, 1 = a^0. When the check succeeds, the integer is the true answer, so the solver reports it as exact instead of as 4.000000. The shortcut only fires for integer bases and results; fractional cases like 4^x = 2 still solve correctly by logarithms, giving x = 0.500000.
Base 1 never changes: 1^x = 1 for every x, so 1^x = b has no solution or infinitely many, and log base 1 is undefined. A negative base breaks down for fractional exponents, since (-2)^0.5 is not a real number. The solver rejects both with The base must be greater than zero or The base cannot be 1.
Yes. A positive base raised to any real power is always positive, so if b is 0 or negative the equation has no real solution. The solver stops with For a^x = b, b must be greater than zero instead of returning a number. The same rule applies to e^x = b and to the value inside log_a(b).
ln is the logarithm with base e, about 2.71828, while log_a lets you choose the base. The solver treats them as one family: e^x = b and ln(x) = b get their own modes because they appear constantly in growth, decay and calculus, and every other base goes through the change of base formula ln(b) / ln(a).
No, it changes only how the answer, the steps and the check are displayed: 2, 4, 6 or 8 decimal places. The underlying value is computed at full precision either way, which is why 2^x = 10 shows 3.32 at 2 decimals and 3.321928 at 6. Exact answers are reported as exact at every setting.
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References
- OpenStax, Rice University. College Algebra 2e, Section 6.6: Exponential and Logarithmic Equations. Accessed October 21, 2026.
- OpenStax, Rice University. College Algebra 2e, Section 6.5: Logarithmic Properties. Accessed October 21, 2026.
- OpenStax, Rice University. College Algebra 2e, Section 6.3: Logarithmic Functions. Accessed October 21, 2026.
- National Institute of Standards and Technology. Digital Library of Mathematical Functions, Section 4.2: Definitions of the Logarithm. Accessed October 21, 2026.
- The LibreTexts Libraries. College Algebra (OpenStax), Section 6.2: Exponential Functions. Accessed October 21, 2026.
- The LibreTexts Libraries. Intermediate Algebra, Section 12.5: Solve Exponential and Logarithmic Equations. Accessed October 21, 2026.