Exponential Growth (Part 2)

Exponential Growth (Part 2)

How to find the amount of time for the exponential growth (compounding yearly, continuously) 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 growth formula:
A = A0(1 + r)t.

Exponential Growth - Final Value

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

Example 1

$1,000 investment is at a rate of 6% per year, compounded yearly. After how many years will the investment worth more than $1,800? (Assume log 1.8 = 0.255, log 1.06 = 0.025.)

The amount of initial investment is $1,000.

So A0 = 1000.

The amount of final investment is $1,800.

So A = 1800.

The rate is 6% per year,
compounded [yearly].

So r = +0.06/year.

A0 = 1000
A = 1800
r = +0.06/year

Then 1000⋅(1 + 0.06)5 = 1800.

Divide both sides by 1000.

Then 1.06t = 1.8.

Log both sides.
(base: 1.06)

Then t = log1.06 1.8.

Logarithmic form

log1.06 1.8 = (log 1.8)/(log 1.06)

Change the base formula

log 1.8 = 0.255
log 1.06 = 0.025

Then (given) = 0.255/0.025.

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

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

255/25 = 10.2

So t = 10.2.

t = 10.2 years

But t should be a natural number.

So round 10.2 up to the nearest ones:
10.2 → 11.

So t = 11 years.

So the answer is [after 11 years].

Formula: Continuous Compounding

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

Recall the continuous exponential growth formula:
A = A0ert.

Continuous Exponential Growth - Final Value

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

Example 2

$1,000 investment is at a rate of 6% per year, continuously compounded. After how many years will the investment worth more than $1,800? (Assume ln 1.8 = 0.588.)

The amount of initial investment is $1,000.

So A0 = 1000.

The amount of final investment is $1,800.

So A = 1800.

The rate is 6% per year,
compounded [yearly].

So r = +0.06/year.

A0 = 1000
A = 1800
r = +0.06/year

Then 1000⋅e0.06⋅t = 1800.

Divide both sides by 1000.

Then e0.06t = 1.8.

Log both sides.
(base: e)

Then 0.06t = ln 1.8.

Natural logarithm

ln 1.8 = 0.588

Divide both sides by 0.06.

Then t = 0.588/0.06.

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

(= Multiply 100
to both of the numerator and the denominator.)

58.8/6 = 9.8

So t = 9.8.

t = 9.8 years

But t should be a natural number.

So round 9.8 up to the nearest ones:
9.8 → 10.

So t = 10 years.

So the answer is [after 10 years].

Compare the answer 9.8 years (continuously)
to the answers of the last example:
10.2 years (yearly).

As you can see,
by compounding the investment continuously,
the amount of time to reach $1,800 decreases.

This is true because
by compounding continuously,
the investment grows faster.