4.6B The Natural Base, e
Objectives:
F.LE.4: For exponential models, express as a logarithm the solution to abct = d where a, c, and d
are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology.
F.IF.7e: Graph exponential and logarithmic functions, showing intercepts and end behavior . . .
F.IF.5: Relate the domain of a function to its graph and, where applicable, to the quantitative
relationship it describes.
Anticipatory Set:
All the properties of logarithms also apply to natural logarithms.
1. ln e = 1 (e1 = e)
2. ln 1 = 0 (e0 = 1)
3. Inverse Property: e ln x  x
4. Product Property
ln a + ln b = ln (ab)
5. Quotient Property
ln a – ln b = ln (a/b)
6. Power Property
ln an = n ln a
Open the book to page 276 and read example 2.
Example: Simplify.
1. ln e0.15t
2. e 3 ln  x 1 
3
e ln  x 1 
0.15t ln e
0.15t
(x + 1)3
White Board Activity:
Practice: Simplify.
1. ln e3.2
3.2 ln e
3.2
2. e2 ln x
elnx
x2
2
3. ln e2x + ln ex
ln (e2x ∙ 3x) = ln e3x
3x ln e = 3x
3. ln e(x + 4y)
(x + 4y) ln e
x + 4y
The most common application of e is the continued compounding of interest.
A = Pert
A is the total amount
P is the principal
r is the annual interest rate (as a decimal)
t is the time in years.
Open the book to page 276 and read example 3.
Example: On the day you were born, your grandmother put $1000 in an investment which pays
5.375% yearly compounded continuously. On your 18th birthday you will get the
proceeds of the investment as a graduation gift. How much will you receive?
A = 1000e(.05375)(18) = $2,631.36
White Board Activity:
Practice: What is the total amount for an investment of $100 invested at 3.5% for 8 years and
compounded continuously?
A = 100e(.035)(8) = $132.31
Another common application of e is the half-life of a substance.
Half-life is the time it takes for half of a substance to breakdown or convert to another substance
during the process of decay.
Natural decay is modeled by
N(t) = N0e-kt
N(t) is the amount remaining after time t
N0 is the initial amount at time t = 0
k is the decay constant
t is the time
Example: Plutonium-239 has a half-life of 24,110 years. How long does it take for a 1 g sample of Pu239 to decay to 0.1g?
Step 1: Find the decay constant.
N(t) = N0e-kt
Choose any value for No, then N(t) = ½ that value.
Let N0 = 1 then N(t) = ½, t = 24110
½ = 1e-k(24110)
Take the logarithm of both sides.
ln ½ = ln e-k(24110)
Apply the Power Property.
ln ½ = -24110k ln e
ln e = 1
ln ½ = -24110k
Solve for k.
k = ln ½ /-24110 = 0.000029
Step 2: Write the decay function.
N(t) = N0e0.000029t
Step 3: Solve the problem: N0 = 1 g and N(t) = 0.1 g
0.1 = 1e-0.000029t
ln 0.1 = ln e-0.000029t
ln 0.1 = -0.000029t ln e
ln 0.1 = -0.000029t
t = ln 0.1/-0.000029 = about 80,000 years
White Board Activity:
Practice: Determine how long it will take for 650 mg of a sample of chromium-51, which has a half-life
of about 28 days, to decay to 200 mg.
Step 1: Find the decay constant.
½ = 1e-k(28)
ln ½ = ln e-k(28)
ln ½ = -k(28) ln e
ln ½ = -28k
k = ln ½ /-28 = 0.025
Step 2: Write the decay function.
N(t) = N0e-0.025t
Step 3: Solve the problem: N0 = ? and N(t) = ?
N0 = 650 and N(t) = 200
200 = 650e-0.025t
200/650 = e-0.025t
ln (200/650) = ln e-0.025t
ln (200/650) = -0.025t ln e
ln (200/650) = -0.025t
t = ln (200/650) / -0.025 = about 47 days
Assessment:
Question student pairs.
Independent Practice:
Text: pgs. 278 – 279 prob. 6 – 12, 17 – 22, 25, 30 – 36.
For a Grade:
Text: pgs. 278 – 279 prob. 10, 12, 18, 22, 30.