Quick answer

If ln(x) = y, then x = e^y = exp(y).

Formula

  • e ≈ 2.71828
  • exp(y) = e^y
  • Inverse pair: ln and exp

Introduction

Natural antilog is the partner of the natural logarithm. Together they describe continuous growth, decay, and rates that change proportionally with the current amount.

Courses label the same operation as e^x, exp, or "antilog with base e." The mathematics is one idea even when the calculator keys look different.

On the home Antilog Calculator, enter y and set base to e (or a decimal approximation) to mirror ln problems from class.

Euler's number and exp

e is the base for natural exponentials. The function exp(y) is defined as e^y, so exp is not a separate formula family.

Calculators intermix ln, e^x, and exp depending on brand; all target the same mathematics when the problem used a natural log.

Engineering formulas for capacitors, RC circuits, and continuous interest use this pairing because many models differentiate cleanly in the ln domain.

If you are comparing when to use e versus 10 in the same week of homework, read common antilog bases for a side-by-side table of notation cues.

Formula applications

ln(x) = y → x = e^y
Growth: A = P e^(rt)
Decay: A = P e^(-kt)

When a model gives ln(A) = k, antilog thinking recovers A with e^k before you report a linear-scale amount.

The step-by-step substitution pattern matches the custom-base method in how to calculate an antilog; only the base digit changes to e.

Natural antilog steps

  1. Confirm natural log. Look for ln, log_e, or wording that says natural logarithm. If the problem used log without ln, check whether your course means base 10 instead.
  2. Apply e^y. Use the exp key or e^y on the calculator. Enter y exactly as given, including negative exponents for decay.
  3. Estimate if needed. e^1 ≈ 2.7, e^2 ≈ 7.4, e^3 ≈ 20. Estimates catch swapped digits before you hand in work.
  4. Verify with ln. Take ln of your result. You should return to the starting y within rounding tolerance.

Engineering-style checks

ln(20) ≈ 3.0 means e^3 ≈ 20. The estimate is close enough for lab notebooks before precise calculator entry.

If ln(population ratio) = 0.05, the multiplier is e^0.05 ≈ 1.051, about a 5.1% increase over the interval.

Continuous interest A = Pe^(rt) is the same antilog idea: solve for P or A by exponentiating when the equation is in ln form.