Exponential Growth (Part 2)

Exponential Growth (Part 2)

How to solve exponential growth 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 growth (part 1)

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.)

Write the given values to use.

Put the given values into the formula.

1000⋅(1 + 0.06)t = 1800

Log both sides. (base: 1.06)

Logarithmic form

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

Change the base formula

log 1.8 = 0.255, log 1.06 = 0.025
(given conditions)

t = 10.2 years

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

So round 10.2 up to the nearest ones:
t = 11 years.

This means after 11 years,
the investment worth more than $1,800.

Formula: Continuous Compounding

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 compounding formula:
A = A0ert

Exponential growth (part 1)

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.)

Write the given values to use.

Put the given values into the formula.

1000⋅e0.06⋅t = 1800

Log both sides. (base: e)

Natural logarithms

ln 1.8 = 0.588
(given condition)

t = 9.8 years

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

So round 9.8 up to the nearest ones:
t = 10 years.

This means after 10 years,
the investment worth more than $1,800.

As you can see,
the answer is greater than
the previous example's answer:

the investment is maximized
by continuous compounding.