Solve for Exponents Calculator (bⁿ = y)

Solve b^n = y for the unknown exponent n given any positive base (other than 1) and positive result, using logarithms, with full steps shown.

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Solving an equation for its exponent

Given an equation of the form bn=yb^n = y — a known base bb and a known result yy — this calculator solves for the unknown exponent nn. The tool for isolating an exponent is the logarithm:

bn=y    n=logbyb^n = y \implies n = \log_b y

For example, 2n=82^n = 8 has the solution n=log28=3n = \log_2 8 = 3, because 23=82^3 = 8. The exponent you are solving for is the logarithm — there is no separate technique for "solving for an exponent" beyond recognizing that a logarithm already does exactly this.

Tip: If you are more used to the log notation logbx\log_b x directly, see the Logarithm calculator — it is the same math, just written the other way around.

Why the logarithm is the right tool

Exponentiation and logarithms are inverse operations, the same way multiplication and division undo each other. Three numbers — the base bb, the exponent nn, and the result yy — are related by both:

KnownUnknownSolve with
bb, nnyyExponentiation: y=bny = b^n
bb, yynnLogarithm: n=logbyn = \log_b y

Because a logarithm undoes exponentiation, plugging the solved exponent back in always recovers the original result: blogby=yb^{\log_b y} = y. This calculator's "The defining identity" section shows this plugged-in check for every answer.

Why the base and result have restrictions

The equation bn=yb^n = y only has a real solution for nn when the base bb satisfies b>0b > 0 and b1b \neq 1, and the result y>0y > 0:

  • b0b \leq 0 is invalid. Raising a negative base to a non-whole exponent (which most solutions are) is not a real number.
  • b=1b = 1 is invalid. 1n=11^n = 1 for every nn, so there is no unique exponent that solves 1n=y1^n = y unless yy is already 1 — and even then, every nn works, so the equation is left undefined.
  • y0y \leq 0 is invalid. A positive base raised to any real power always stays positive, so no real nn makes bnb^n zero or negative.

This calculator reports any of those combinations as out of range instead of guessing at a misleading number.

Exact exponents vs. decimal approximations

When the result is an exact power of the base — like 2n=82^n = 8 or 10n=100010^n = 1000 — the exponent is a whole number, computed exactly. Most results are not exact powers, so the exponent is an irrational decimal, rounded here to 6 decimal places.

Example 1 (exact): 2n=8    n=32^n = 8 \implies n = 3, since 23=82^3 = 8.

Example 2 (exact): 10n=1000    n=310^n = 1000 \implies n = 3, since 103=100010^3 = 1000.

Example 3 (exact): 3n=81    n=43^n = 81 \implies n = 4, since 34=813^4 = 81.

Example 4 (decimal approximation): 2n=10    n3.3219282^n = 10 \implies n \approx 3.321928, since 23.321928102^{3.321928} \approx 10.

Where solving for an exponent shows up

Solving for an unknown exponent is how you answer "how many times / how long / how many periods" questions in exponential growth and decay — compound interest asking how many years to double, radioactive decay asking how many half-lives have passed, or population growth asking how many generations to reach a target size. In every case the growth rate is the base, the target is the result, and the unknown number of steps is the exponent you solve for with a logarithm. For the reverse question — raising a known base to a known power — see the Exponents calculator and the Fractional Exponents calculator.

Frequently asked questions

How do you solve for an exponent?
To solve bn=yb^n = y for the exponent nn, take the logarithm of both sides in base bb: n=logbyn = \log_b y. For example, solving 2n=82^n = 8 gives n=log28=3n = \log_2 8 = 3, because 23=82^3 = 8.
Why does taking a logarithm solve for the exponent?
A logarithm is defined as the inverse of exponentiation: logby\log_b y is, by definition, the power you raise bb to in order to get yy. So bn=yb^n = y and n=logbyn = \log_b y are two ways of writing the exact same relationship — the logarithm just isolates nn.
Why must the base be positive and not equal to 1?
If the base b0b \leq 0, raising it to most real powers is not a real number (a negative base to a fractional power is undefined over the reals). If b=1b = 1, then 1n=11^n = 1 for every nn, so there is no single exponent that solves the equation — it is left undefined rather than guessed at.
Why must the result be positive?
A positive base raised to any real power always stays positive, so there is no real exponent nn that makes bnb^n equal zero or a negative number when b>0b > 0. If the result you are solving for is 0\leq 0, the equation has no real solution.
Is the exponent always a whole number?
No. The exponent is only a whole number when the result is an exact power of the base, like 2n=82^n = 8 giving n=3n = 3. Most equations do not work out this cleanly — solving 2n=102^n = 10 gives an irrational decimal, n3.321928n \approx 3.321928, which this calculator rounds to 6 decimal places.
How is this different from the Logarithm calculator?
It is the same computation, framed differently. The [Logarithm calculator](/en/calculators/algebra/logarithm) is written as logbx\log_b x directly; this calculator is written as the equation bn=yb^n = y that you are solving for nn. Both compute n=logbyn = \log_b y under the hood.

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