Base ‘e’ and Natural Logarithms
Example 1
Evaluate Natural Base Expressions
Use a calculator to evaluate each expression to four decimal places.
a. e2.5
KEYSTROKES: 2nd [ex] 2.5 ENTER
12.182493961
e 2.5 12.1825
–4.6
x
b. e
KEYSTROKES: 2nd [e ] –4.6 ENTER
0.010051836
e 4.6 0.0101
Example 2
Evaluate Natural Logarithmic Expressions
Use a calculator to evaluate each expression to four decimal places.
a. ln 15
KEYSTROKES: LN 15 ENTER
2.708050201
ln 15 2.7081
b. ln 0.75
KEYSTROKES: LN 0.75 ENTER
–0.287682072
ln 0.75 -0.2877
Example 3
Write Equivalent Expressions
Write an equivalent exponential or logarithmic equation.
a. e–x = 2
b ln x = 0.35
e–x = 2
loge 2 = -x
ln x = 0.35
ln 2 = -x
loge x = 0.35
x = e0.35
Example 4
Solve Base e Equations
Solve –2e + 10 = 6.
5x
–2e5x + 10 = 6
–2e5x = –4
e5x = 2
ln e5x = ln 2
5x = ln 2
x=
1
5
(ln 2)
x ≈ 0.1386
Original equation
Subtract 10 from each side.
Divide each side by –2.
Property of Equality for Logarithms
Inverse Property of Exponents and Logarithms
1
Multiply each side by .
5
Use a calculator.
The solution is about 0.1386.
CHECK
Substitute this value into the original equation.
–2e5x + 10 = 6
–2e
5(0.1386)
+ 10
?
?
6
–3.9994 + 10 6
6.0006 ≈ 6 
Original equation
Substitute 0.1386 for x.
Simplify.
Simplify.
Example 5
Solve Base e Inequalities
Savings Suppose you deposit $5000 in an account paying 4% annual interest, compounded
continuously.
a. What is the balance after 8 years?
A = Pe rt
= 5000e(0.04 · 8)
= 5000e 0.32
6885.64
Continuous compounding formula
Replace P with 5000, r with 0.04, and t with 8.
Simplify.
Use a calculator.
The balance after 8 years would be $6885.64.
CHECK
If the account was earning simple interest, the formula for the interest would be I = Prt.
In that case, the interest would be I = 5000(0.04)(8) or $1600. Continuously
compounded interest should be greater than simple interest at the same rate. Thus the
solution $6885.64 is reasonable.
b. How long will it take for the balance in your account to reach at least $10,000?
WORDS
The balance is at least $10,000.
VARIABLE
Let A represent the amount in the account.
INEQUALITY A
10,000
5000e 0.04t
e 0.04t
ln e 0.04t
0.04t
t
t
0.04t
10,000
2
ln 2
ln 2
Let A = 5000e
.
Divide both sides by 5000.
Property of Equality for Logarithms.
Inverse Property of Exponents and Logarithms
ln 2
0.04
Divide each side by 0.04.
17.33
Use a calculator.
It will take at least 17.33 years for the balance to reach $10,000.
Example 6
Solve Natural Log Equations and Inequalities
Solve each equation or inequality.
a. ln (6x – 3) + 3 = 10
ln (6x – 3) + 3 = 10
ln (6x – 3) = 7
eln (6x – 3) = e7
6x – 3 = e7
6x = e7 + 3
e
x=
7
Original equation
Subtract 3 from each side.
Write each side using exponents and base e.
Inverse Property of Exponents and Logarithms
Add 3 to each side.
3
6
Divide each side by 6.
x ≈ 183.2722
Use a calculator.
The solution is about 183.2722. Check this solution using substitution or by graphing.
b. ln (3x + 2) < 5
ln (3x + 2) < 5
eln (3x + 2) < e5
3x + 2 < e5
3x < e5 – 2
x<
e
5
Original inequality
Write each side using exponents and base e.
Inverse Property of Exponents and Logarithms
Subtract 2 from each side.
– 2
3
x < 48.8044
Divide each side by 3.
Use a calculator.
Recall that (3x + 2) must be greater than 0 in order for ln (3x + 2) to be defined. Therefore, you must
solve 3x + 2 > 0.
3x + 2 > 0
3x > –2
x>–
Inequality to solve
Subtract 2 from each side.
2
Divide each side by 3.
3
The solution is –
CHECK
2
3
< x < 48.8044.
Test values to the left and right of this interval and in the interval to check the solution.
Test x = –0.75.
ln (3x + 2) < 5
ln [3(–0.75) + 2]
ln (–0.25)
?
?
Test x = 0.
ln (3x + 2) < 5
5
ln [3(0) + 2]
5
ln (2)
ln (–0.25) is not defined.
?
?
Test x = 50.
ln (3x + 2) < 5
5
ln [3(50) + 2]
5
ln (152)
0.6931 < 5 
?
?
5
5
5.0239 > 5