Solve logarithmic equations in seconds — find the missing base, argument, or result of log_b(x) = y with domain checks and step-by-step working. Free, fast, and mobile-friendly.
Home » Math Calculators » Logarithm Equation Calculator
Logarithms turn multiplication into addition and exponentiation into multiplication — but solving an equation like log₂(x) = 5 still stops many students cold. Which part do you exponentiate? What if the base is the unknown? Our free Logarithm Equation Calculator solves all three forms of log_b(x) = y: give it any two of the base, the argument, and the result, and it finds the third with full step-by-step working and automatic domain checks.
The key insight is small: log_b(x) = y is just another way of writing b^y = x. Every logarithm equation is an exponential equation in disguise, and converting between the two forms is the entire skill. The calculator performs that conversion visibly — exponentiating to find x, applying change-of-base to find y, or taking roots to find b — so each solution teaches the method while delivering the answer.
Logarithmic equations model real phenomena: the pH scale, the Richter scale, decibel levels, compound interest time-to-double, and algorithm complexity are all log equations waiting to be solved. Whether you are doing homework or sizing a real system, the calculator handles the algebra and flags the domain traps (non-positive arguments, base 1) that textbooks love to test.
A logarithm is an exponent: log_b(x) is the power you raise b to in order to get x. So log₂(8) = 3 because 2³ = 8; log₁₀(1000) = 3 because 10³ = 1000; and log₅(1) = 0 because 5⁰ = 1. The notation log_b(x) = y and the equation b^y = x say the identical thing — converting fluently between them is the master key to every log problem.
Three domain rules govern all real-valued logarithms, and they all come from the exponential form. Rule 1: x > 0. Since b^y is always positive for positive b, no exponent produces zero or a negative — log₂(−8) has no real answer. Rule 2: b > 0 and b ≠ 1. A negative base gives chaotic sign flips, base 0 is degenerate, and base 1 is useless because 1^y = 1 always (log₁(5) would ask “1 to what power is 5?” — impossible). The calculator enforces all three rules with specific messages.
The change-of-base formula, log_b(x) = ln(x)/ln(b), is how every calculator actually computes logarithms: hardware knows ln (base e) natively, and everything else is a division. It also reveals a deep unity — all logarithm functions are proportional to each other, differing only by a constant factor 1/ln(b). That is why “bits” (base 2), “nats” (base e), and “decibels” (base 10, scaled) measure the same information in different units.
Logarithms invert exponentials the way subtraction inverts addition, which makes them the natural tool for “how long until…?” questions. If a population doubles every 7 years (P = P₀·2^(t/7)), the doubling time from any starting point is one line of logs. Anytime a quantity grows multiplicatively and you need the time or count, a logarithm equation is hiding in the problem.
log_b(x) = y ⟺ b^y = x | log_b(x) = ln(x)/ln(b) | b = x^(1/y) when solving for the base
Core log laws: log_b(mn) = log_b(m) + log_b(n); log_b(m/n) = log_b(m) − log_b(n); log_b(m^k) = k·log_b(m). Domain: b > 0, b ≠ 1, x > 0.
Convert to exponential form: x = 2⁵ = 32. Check: 2⁵ = 32. ✓ The domain is automatic here — a power of a positive base is always positive, so no extraneous solutions can sneak in when solving for x.
Ask “3 to what power is 81?” Since 3⁴ = 81, y = 4. Via change-of-base: ln(81)/ln(3) ≈ 4.3944/1.0986 ≈ 4. When the argument is an exact power of the base, look for the integer first — it is faster and exact.
Convert: b³ = 125, so b = 125^(1/3) = 5. Check: 5³ = 125. ✓ Note the domain subtlety: b³ = 125 also has complex roots, but the real logarithm demands b > 0, b ≠ 1 — only 5 qualifies.
10 is not a power of 2, so use change-of-base: ln(10)/ln(2) ≈ 2.3026/0.6931 ≈ 3.3219. Sanity check: 2³ = 8 < 10 < 16 = 2⁴, so the answer must lie between 3 and 4 — and 3.3219 does. This bracketing habit catches most arithmetic slips.
How long until an investment doubles at 7% annual compound interest? The equation is (1.07)^t = 2, so t = log(2)/log(1.07) ≈ 0.6931/0.0677 ≈ 10.24 years — the Rule of 72 (72/7 ≈ 10.3) as an exact computation. Set the calculator to solve for y with b = 1.07, x = 2, and it walks through the change-of-base step.
The pH scale is a base-10 log equation: pH = −log₁₀[H⁺]. If a solution’s hydrogen-ion concentration is 3.2 × 10⁻⁵ mol/L, then pH = −log₁₀(3.2 × 10⁻⁵) ≈ 4.49 — mildly acidic, like tomato juice. Solving backward (find [H⁺] from pH 4.49) is the “solve for x” mode: x = 10^(−4.49).
Computer scientists ask “how many bits to represent N values?” — that is log₂(N). A million values need log₂(1,000,000) ≈ 19.93, so 20 bits. And the Richter scale: a magnitude-7 quake releases 10^(1.5×7) energy units versus 10^(1.5×5) for magnitude 5 — the ratio is 10^3 = 1000×. Every one of these is a logarithm equation with real stakes.
The most dangerous log equations are the ones that look solved but harbor extraneous roots. Consider log₂(x) + log₂(x − 3) = 5. Combining: log₂(x(x−3)) = 5, so x(x−3) = 32, giving x = 8 or x = −5. But x = −5 makes both original logarithms undefined (negative arguments) — it is extraneous, born the moment we combined the logs, because log₂(x) + log₂(x−3) = log₂(x(x−3)) is only valid when x > 0 and x − 3 > 0 individually.
The general principle: every log law (product, quotient, power) carries domain fine print, and applying the law can widen the domain, admitting impostor solutions. The defense is a final domain audit: substitute each candidate back into the original equation and confirm every logarithm’s argument is positive and every base is valid. The calculator runs these checks on its own solutions automatically.
This is also why textbooks drill “state the domain first.” For log₂(x) + log₂(x−3), the domain is x > 3, which kills x = −5 before it is even born. Five seconds of domain analysis saves five minutes of chasing ghosts — professional habit, not pedantry.
x = 4^2.5 = 4² × 4^0.5 = 16 × 2 = 32. Fractional logarithms are painless once you split the exponent: integer part by repeated multiplication, fractional part by roots.
b⁴ = 10000, so b = 10000^(1/4) = 10. Check: 10⁴ = 10000. ✓ Recognizing 10000 = 10⁴ makes this instant — powers of 10 are the log world’s home turf.
A unifying view: the three solution modes are the three inverse operations of exponentiation. Fixing b and y and solving for x is plain exponentiation (x = b^y). Fixing b and x and solving for y is the logarithm proper (y = ln x / ln b). Fixing x and y and solving for b is root extraction (b = x^(1/y)). The calculator’s mode dropdown is really a tour of the inverse-operation family — master all three and no exponential equation can surprise you.
Solve log₅(x) = 3 for x, then verify by computing 5³. Next solve log₅(200) for y and bracket your answer between consecutive integers first.
Find the base b such that log_b(81) = 4. (Hint: b⁴ = 81 — and remember b > 0, b ≠ 1.)
A fintech startup’s user base grows 12% month-over-month, and the founders ask: when do we 10×? The equation is 1.12^t = 10, so t = log(10)/log(1.12) ≈ 2.3026/0.1133 ≈ 20.3 months. That single number drives hiring plans, server budgets, and fundraising timelines — and it is a logarithm equation solved in the calculator’s “solve for y” mode (b = 1.12, x = 10).
The CFO then asks the inverse question: what monthly growth rate 10×s us in exactly 12 months? Now the base is unknown: b^12 = 10 gives b = 10^(1/12) ≈ 1.2115, i.e. 21.15% monthly. Same calculator, “solve for b” mode — the two questions are mirror images, and fluency in both directions is what turns a formula into a business tool.
Notice what never appeared: trial and error. Each question was one exact computation, because the team reached for logarithms instead of guessing. That is the real dividend of log fluency — replacing iteration with insight.
The Logarithm Equation Calculator solves log_b(x) = y for any one unknown: it exponentiates (x = b^y) to find the argument, uses change-of-base (y = ln x / ln b) to find the logarithm, and takes roots (b = x^(1/y)) to find the base — enforcing the domain rules b > 0, b ≠ 1, x > 0 and showing each step.
Convert to exponential form: x = 2⁵ = 32. In general, log_b(x) = y becomes x = b^y — exponentiate both sides and evaluate.
Rewrite as b³ = 125, then take the cube root: b = 125^(1/3) = 5. Remember the base must be positive and not equal to 1.
log_b(x) = ln(x) / ln(b) — it converts any logarithm into natural logs (or common logs), which is how calculators actually evaluate arbitrary bases.
Because 1 raised to any power is 1, so log₁(x) would ask “1 to what power equals x?” — impossible for x ≠ 1 and ambiguous for x = 1. Base 1 logarithms are undefined.
Not for real-valued logarithms: b^y is always positive for b > 0, so no exponent produces a negative number or zero. log₂(−8) has no real answer (complex logarithms exist but are beyond this calculator).
Solutions introduced when applying log laws widened the domain — e.g. combining log(x) + log(x−3) admits x = −5, which makes the original logs undefined. Always check candidates in the original equation.
Set (growth factor)^t = 2 and solve: t = log(2)/log(1 + rate). At 7% annual growth, t = ln 2 / ln 1.07 ≈ 10.24 years — the exact form of the Rule of 72.
They are the same function with different bases: log usually means base 10, ln means base e (≈ 2.718), and log₂ means base 2. Change-of-base converts freely: they differ only by a constant factor.
Cite this page as: Logarithm Equation Calculator. Sigma Calculator. Retrieved October 9, 2026, from https://sigmacalculator.net/math-calculators/logarithm-equation-calculator/
Last updated: October 9, 2026