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Using Exponential & Logarithmic Functions

Exponential and logarithmic functions model the same processes from two directions: exponentials find an amount after time, and logs find the time to reach an amount.

Continuous interest with A=PertA = Pe^{rt} is the classic application. Evaluate it directly for a future value, or take a natural log to solve for time.

Evaluating an exponential model

Plug the principal, rate, and time into A=PertA = Pe^{rt} and compute. Use the decimal rate and the calculator's ee key.

The result is the amount after continuous growth over that time.

Logs solve for the exponent

When time is the unknown, isolate the exponential and take the natural log of both sides. ln\ln pulls the exponent down front.

This turns 'how long until it doubles?' into a solvable equation, since lnert=rt\ln e^{rt} = rt.

Worked examples

Example 1: future value

$2000 grows at 3% compounded continuously. Using A=PertA = Pe^{rt}, find its value after 5 years.

SubstituteA=2000e0.035A = 2000 e^{0.03 \cdot 5}
Simplify the exponent2000e0.152000 e^{0.15}
Evaluate2324\approx 2324

Answer: About $2324

Example 2: solving for time

Which operation isolates tt in A=PertA = Pe^{rt}?

Divide by P, then take lnln(AP)=rt\ln\left(\dfrac{A}{P}\right) = rt

Answer: Take the natural log

Try one yourself

Common questions

When do I use a log instead of an exponential?

Use the exponential to find an amount after known time; use a logarithm when the time (the exponent) is what you are solving for.

Why the natural log with erte^{rt}?

ln\ln is the inverse of exe^x, so lnert=rt\ln e^{rt} = rt — it cleanly brings the exponent down to solve for tt.

What rate goes in the formula?

The decimal form: 3% becomes 0.030.03.

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