Math
EXP function: model continuous compounding and exponential decay
Raise Euler's constant e (2.71828...) to any power with EXP to calculate continuous growth P*e^(r*t) and exponential decay.
Continuous growth vs discrete annual compounding
EXP(number) computes e^number, where e ≈ 2.718281828 is the base of natural logarithms. In finance, physics, and biology, continuous growth over time t at rate r follows P * EXP(r * t).
On a 1,000 principal at 6% for 5 years, annual compounding 1000*(1+0.06)^5 grows to 1,338.23, whereas continuous compounding 1000*EXP(0.06*5) = 1000*e^0.3 grows to 1,349.86.
| Initial Value P (A) | Rate r (B) | Time t (C) | P * EXP(r * t) |
|---|---|---|---|
| 1000 | 0.06 | 5 | 1349.86 (continuous growth) |
| 1000 | 0.06 | 0 | 1000 (EXP(0) = 1) |
| 1000 | -0.15 | 4 | 548.81 (exponential decay) |
Continuous compounding and decay formula
Pass a negative rate r (such as -0.15 per hour for a temperature difference from ambient in a constant-rate cooling model or radioactive/chemical decay) and EXP smoothly decays toward zero without ever turning negative.
=ROUND(A2 * EXP(B2 * C2), 2)Zero exponent and inverse relationship with LN
At t = 0, EXP(0) equals exact 1 so initial principal is returned unchanged. EXP(1) returns Euler's constant e (rounded to 2.718282).
=EXP(0)Try the example
Use the original values or change them to test your understanding.