Exponential Decay (Part 2)

Exponential Decay (Part 2)

How to solve exponential decay problems using logarithmics: formulas, examples, and their solutions.

Formula

A = A_0(1 - r)^t. A: Final Value, A_0: Initial Value, r: Rate of Change (per Period), t: Number of Period

Recall the exponential growth formula:
A = A0(1 - r)t

Exponential decay (part 1)

By using the logarithm,
now you can find the value of t.

Example 1

The weight of a new eraser is 100g. If it is decreasing at a rate of 14% per week, after how many weeks will the weight become less than 10g? (Assume log 0.86 = -0.066.)

Write the given values to use.

Put the given values into the formula.

100⋅(1 - 0.14)t = 10

Log both sides. (base: 0.86)

Logarithmic form

To use the given condition,
change the base to 10.

Change the base formula

log (1/10) = -1
(See the right formula: logb (1/b) = -1)

log 0.86 = -0.066
(given condition)

t = 15.xx weeks

But the question says 'how many weeks'.
So t is a natural number.

So round 15.xx up to the nearest ones:
t = 16 weeks.

This means after 16 weeks,
the weight of the eraser becomes less than 10g.

Formula: Continuous Decay

A = A_0*e^rt. A: Final Value, A_0: Initial Value, r: Rate of Change (per Period), t: Number of Period

Recall the continuous decay formula:
A = A0e-rt

Exponential decay (part 1)

By using the logarithm,
you can also find t from this formula.

Example 2

A radioactive substance is continuously decreasing at a rate of 5% per minute. Find the half-life of the substance. (Assume ln 2 = 0.69.)

Write the given values to use.

The half-life is the amount of time that takes
to change from A0 to (1/2)A0.

So A = (1/2)A0.

Put the given values into the formula.

A0e-0.05⋅t = (1/2)A0

Log both sides. (base: e)

Natural logarithms

ln (1/2) = -ln 2
(See the right formula: logb (1/c) = -logb c)

Logarithms of powers

ln 2 = 0.69
(given condition)

This means
the amount of quantity (weight, concentration, etc.)
will be half in every 69/5 minutes (= 13.8 minutes).