Exponential Decay (Part 2)

Exponential Decay (Part 2)

How to find the amount of time for the exponential decay by using logarithm: formulas, examples, and their solutions.

Formula

A = A0(1 + r)^t. A: Final value, A0: Initial value, -r: Rate of change (per period), t: Number of period

Recall the exponential change formula:
A = A0(1 + r)t.

Exponential Decay - Final Value

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.)

The initial weight is 100g.

So A0 = 100.

The final weight is 10g.

So A = 10.

The rate is 14% per week,
decreasing [weekly].

So -r = -0.14/week.

A0 = 100
A = 10
-r = -0.14/week

So 100⋅(1 - 0.14)t = 10.

Divide both sides by 100.
And 1 - 0.14 = 0.86.

Then 0.86t = 1/10.

Log both sides.
(base: 0.86)

Then Then t = log0.86 1/10.

Logarithmic form

log0.86 1/10 = (log 1/10)/(log 0.86)

Change the base formula

1/10 = 10-1

Negative exponent

log 10-1 = -1 log 10

Logarithm of a power

log 10 = 1

Logarithm of the base

log 0.86 = -0.066

Cancel the (-) signs
on both of the numerator and the denominator.

Then (given) = 1.000/0.066.

Move the decimal points of both numbers
3 digits to the right.

(= Multiply 1000
to both of the numerator and the denominator.)

Reduce 1000 to 500
and reduce 66 to 33.

500/33 = 15.xx

So t = 15.xx.

t = 15.xx weeks.

But t should be a natural number.

So round 15.xx up to the nearest ones:
15.xx → 16.

So t = 16 weeks.

So the answer is [after 16 weeks].

Formula: Continuous Decay

A = A0*e^rt. A: Final value, A0: Initial value, -r: Rate of change (per period), t: Number of period

Recall the continuous exponential change formula:
A = A0ert.

Continuous Exponential Decay - Final Value

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.)

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

So set A0 = A0
and A = A0/2.

The rate is 5% per minute,
decreasing [minutely].

So -r = -0.05/minute.

A0 = A0
A = A0/2
-r = -0.05/minute

Then A0e-0.05⋅t = A0/2.

Divide both sides by A0.

Then e-0.05t = 1/2.

Natural log both sides.

Then -0.05t = ln (1/2).

Natural logarithm

1/2 = 2-1

ln 2-1 = -1 ln 2

Logarithm of a power

ln 2 = 0.69

-1⋅0.69 = -0.69

Cancel the (-) signs in both sides.

Move the decimal points of both sides
2 digits to the right.

(= Multiply 100 to both sides.)

Divide both sides by 5.

Then t = 69/5.

Then the half-life of the substance is
69/5 minutes.
(= 13.8 minutes)

When finding the half-life,
the answer don't have to be a natural number.

So you don't have to
round the answer up to the nearest ones.