7-6
The Natural Base, e
The natural logarithm, “ln,” has the same
properties as log base 10 and logarithms with
other bases.
The domain of f(x) = ln x is D: (0, ∞).
The range of f(x) = ln x is R: (-∞, ∞).
The natural logarithmic function f(x) = ln x is the
inverse of the natural exponential function f(x) = ex.
Holt Algebra 2
7-6
The Natural Base, e
Example 2: Simplifying Expression with e or ln
Simplify.
A. ln e0.15t
B. e3ln(x +1)
ln e0.15t = 0.15t
C. ln e2x + ln ex
ln e2x + ln ex = 2x + x = 3x
Holt Algebra 2
e3ln(x +1) = (x + 1)3
7-6
The Natural Base, e
Solve the equation.
9 − 𝑙𝑛3𝑥 = 10
−𝑙𝑛3𝑥 = 1
Isolate the ln.
𝑙𝑛3𝑥 = −1
𝑒 𝑙𝑛3𝑥 = 𝑒 −1
Raise each side of the equation to the e power.
3𝑥 = 𝑒 −1
𝑒 −1
𝑥=
3
1
𝑥=
≈ .123
3𝑒
Holt Algebra 2
Solve for x.
7-6
The Natural Base, e
Solve the equation.
𝑒 3𝑥
2+
= 22
2
𝑒 3𝑥
= 20
2
𝑒 3𝑥 = 40
ln⁡(𝑒 3𝑥 ) = 𝑙𝑛40
3𝑥 = 𝑙𝑛40
𝑙𝑛40
𝑥=
≈ 1.230
3
Holt Algebra 2
Isolate the exponential.
Take the ln of both sides.
Solve for x.
7-6
The Natural Base, e
The half-life of a substance is the time it takes for
half of the substance to breakdown or convert to
another substance during the process of decay.
Natural decay is modeled by the function below.
Holt Algebra 2
7-6
The Natural Base, e
Example 4: Science Application
Plutonium-239 (Pu-239) has a half-life of
24,110 years. How long does it take for a 1 g
sample of Pu-239 to decay to 0.1 g?
Step 1 Find the decay constant for plutonium-239.
N(t) = N0e–kt
1 = 1e–k(24,110)
2
Holt Algebra 2
Use the natural decay function.
Substitute 1 for N0 ,24,110 for t,
and 21 for N(t) because half of the
initial quantity will remain.
The Natural Base, e
7-6
Example 4 Continued
ln 21 = ln e–24,110k Simplify and take ln of both sides.
–1
ln 2
1 as 2 –1, and simplify the
Write
= –24,110k
2
right side.
–ln 2 = –24,110k
ln2
k = 24,110
Holt Algebra 2
ln2–1 = –1ln 2 = –ln 2
≈ 0.000029
7-6
The Natural Base, e
Example 4 Continued
Step 2 Write the decay function and solve for t.
N(t) = N0e–0.000029t
0.1 = 1e–0.000029t
Substitute 0.000029 for k.
Substitute 1 for N0 and 0.01 for N(t).
ln 0.1 = ln e–0.000029t
Take ln of both sides.
ln 0.1 = –0.000029t
Simplify.
ln 0.1
t = – 0.000029
≈ 80,000
It takes approximately 80,000 years to decay.
Holt Algebra 2
7-6
The Natural Base, e
Check It Out! Example 4
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 for Chromium-51.
N(t) = N0e–kt
1
2
–k(28)
= 1e
Holt Algebra 2
Use the natural decay function. t.
Substitute 1 for N0 ,28 for t, and 21 for
N(t) because half of the initial quantity
will remain.
The Natural Base, e
7-6
Check It Out! Example 4 Continued
ln 21 = ln e–28k
–1
ln 2
= –28k
–ln 2 = –28k
Simplify and take ln of both sides.
Write 21 as 2 –1 , and simplify the
right side.
ln 2 –1 = –1ln 2 = –ln 2.
k = ln 2 ≈ 0.0247
28
Holt Algebra 2
7-6
The Natural Base, e
Check It Out! Example 4 Continued
Step 2 Write the decay function and solve for t.
N(t) = N0e–0.0247t
Substitute 0.0247 for k.
200 = 650e–0.0247t
Substitute 650 for N0 and 200 for
N(t).
Take ln of both sides.
ln 200 = ln e–0.0247t
650
ln 200 = –0.0247t
650
ln 200
650
t = –0.0247 ≈ 47.7
Simplify.
It takes approximately 47.7 days to decay.
Holt Algebra 2
7-6
The Natural Base, e
HW Natural Log Worksheet on Homeroom
Work through all problems.
#’s 7-9 find intersection using calculator
Holt Algebra 2