LESSON 3–4
Exponential and
Logarithmic Equations
A. Solve 4x + 2 = 16x – 3.
4x + 2 = 16x – 3
Original equation
4x + 2 = (42)x – 3
42 = 16
4x + 2 = 42x – 6
Power of a Power
x + 2 = 2x – 6
One-to-One Property
2 =x–6
Subtract x from each side.
8 =x
Add 6 to each side.
Answer: 8
Solve Exponential Equations Using One-toOne Property
B. Solve
.
Original equation
Power of a Power
One-to-One Property.
Solve 25x + 2 = 54x.
A. 1
B.
C. 2
D. –2
Solve Logarithmic Equations Using One-toOne Property
A. Solve 2 ln x = 18. Round to the nearest
hundredth.
Method 1
Use exponentiation.
2 ln x = 18
Method 2
Original equation
ln x = 9
Divide each side by 2.
eln x = e9
Exponentiate each side.
x = e9
Inverse Property
x ≈ 8103.08
Use a calculator.
Answer: 8103.08
Write in exponential form.
2 ln x = 18
ln x = 9
x = e9
Original equation
Divide each side by 2.
Write in exponential
form.
x ≈ 8103.08
Use a calculator.
Solve Logarithmic Equations Using One-toOne Property
B. Solve 7 – 3 log 10x = 13.
7 – 3 log 10x = 13
–3 log 10x = 6
log 10x = –2
Original equation
Subtract 7 from each side.
Divide each side by –3.
10–2 =10x
Write in exponential form.
10–3 = x
Divide each side by 10.
=x
Answer:
= 10–3.
Solve 2 log2x 3 = 18.
A. 81
B. 27
C. 9
D. 8
A. Solve log2 5 = log2 10 – log2 (x – 4).
log25 = log210 – log2(x – 4)
Original equation
log25 =
Quotient Property
5 =
One-to-One Property
5x – 20 = 10
Multiply each side by
x – 4.5
5x = 30
Add 20 to each side.
x =6
Answer: 6
Divide each side by 5.
Solve Exponential Equations Using One-toOne Property
B. Solve log5 (x2 + x) = log5 20.
log5(x2 + x) = log520
x2 + x = 20
Original equation
One-to-One Property
x2 + x – 20 = 0
Subtract 20 from each
side.
(x – 4)(x + 5) = 0
Factor x 2 + x – 20 into
linear factors.
x = –5 or 4
Answer: –5, 4
Solve for x. Check this
solution.
Solve log315 = log3x + log3(x – 2).
A. 5
B. –3
C. –3, 5
D. no solution
Solve Exponential Equations
A. Solve 3x = 7. Round to the nearest hundredth.
3x = 7
Original equation
log 3x = log 7
Take the common logarithm of
each side.
x log 3 = log 7
x =
Answer:
Power Property
or about 1.77
Divide each side by log 3 and
use a calculator.
1.77
B. Solve e2x + 1 = 8. Round to the nearest hundredth.
e2x + 1 = 8
Original equation
ln e2x + 1 = ln 8
Take the natural logarithm of
each side.
2x + 1 = ln 8
x =
Answer:
Inverse Property
or about 0.54
0.54
Solve for x and use a
calculator.
Solve 4x = 9. Round to the nearest hundredth.
A. 0.63
B. 1.58
C. 2.25
D. 0.44
Solve in Logarithmic Terms
Solve 36x – 3 = 24 – 4x. Round to the nearest
hundredth.
36x – 3 = 24 – 4x
Original equation
ln 36x – 3 = ln 24 – 4x
(6x – 3) ln 3 = (4 – 4x) ln 2
Take the natural
logarithm of each side.
Power Property
6x ln 3 – 3 ln3 = 4 ln 2 – 4x ln 2 Distributive Property
6x ln 3 + 4x ln 2 = 4 ln 2 + 3 ln3
Isolate the variable on
the left side of the
equation.
x(6 ln 3 + 4 ln 2) = 4 ln 2 + 3 ln3
Distributive Property
(could just divide here… hard to type in)
Solve in Logarithmic Terms
x(ln 36 + ln 24) = ln 24 + ln 33
Power Property
x ln [36(24)] = ln [24(33)]
Product Property
x ln 11,664 = ln 432
36(24) = 11,664 and
24(33) = 432
x=
Divide each side by
ln 11,664.
x ≈ 0.65
Use a calculator.
Answer: 0.65
Solve 4x + 2 = 32 – x. Round to the nearest hundredth.
A. 1.29
B. 1.08
C. 0.68
D. –0.23
Solve Exponential Equations in Quadratic
Form
Solve e2x – ex – 2 = 0.
e2x – ex – 2 = 0
Original equation
u2 – u – 2 = 0
Write in quadratic form
by letting u = ex.
(u – 2)(u + 1) = 0
Factor.
u = 2 or u = –1 Zero Product Property
ex = 2
ln ex = ln 2
ex = –1
ln ex = ln (–1)
x = ln 2
or about 0.69
x = ln (–1)
Replace u with ex.
Take the natural
logarithm of each side.
Inverse Property
The only solution is x = ln 2 because ln (–1) is extraneous.
Answer: 0.69
Solve e2x + ex – 12 = 0.
A. ln 3
B. ln 3, ln 4
C. ln 4
D. ln 3, ln (–4)
Solve Logarithmic Equations
Solve log x + log (x – 3) = log 28.
log x + log (x – 3) = log 28
Original equation
log x(x – 3) = log 28
Product Property
log (x 2 – 3x) = log 28
x 2 – 3x = 28
x 2 – 3x – 28 = 0
(x – 7)(x + 4) = 0
x = 7 or x = – 4
Simplify.
One-to-One Property
Subtract 28 from each side.
Factor.
Zero Product Property
The only solution is x = 7 because –4 is an extraneous solution.
Answer: 7
Check for Extraneous Solutions
Solve log (3x – 4) = 1 + log (2x + 3).
log (3x – 4) = 1 + log (2x + 3)
log (3x – 4) – log (2x + 3) = 1
=1
= log 101
= log 10
= 10
3x – 4 = 10(2x + 3)
3x – 4 = 20x + 30
–17x = 34
x = –2
But wait… must check for extraneous solutions
Check for Extraneous Solutions
Check
log (3x – 4) = 1 + log (2x + 3)
log (3(–2) – 4) = 1 + log (2(–2) + 3)
log (–10) = 1 = log (–1)
Since neither log (–10) nor log (–1) is defined, x = –2 is
an extraneous solution.
Answer: no solution
Model Exponential Growth
A. CELL PHONES This table shows the number of
cell phones a new store sold in March and August
of the same year. If the number of phones sold per
month is increasing at an exponential rate, identify
the continuous rate of growth. Then write the
exponential equation to model this situation.
Model Exponential Growth
Let N(t) represent the number of cell phones sold at
the end of t months and assume continuous growth.
Then the initial number N0 is 88 cell phones sold and
the number of cell phones sold N after a time of 5
months, the number of months from March to August,
is 177. Use this information to find the continuous
growth rate k.
N(t) = N0ekt
Exponential Growth Formula
177 = 88e5k
N(5) = 177, N0 = 88, and t = 5
= e5k
Divide each side by 88.
Model Exponential Growth
= ln e5k
Take the natural logarithm of
each side.
= 5k
Inverse Property
=k
Divide each side by 5
0.1398 ≈ k
Use a calculator.
Answer: 13.98%; N(t) = 88e0.1398t
Model Exponential Growth
B. CELL PHONES Use your model to predict the
number of months it will take for the store to sell
500 phones in one month.
N(t) = 88e0.1398t
N(t) = 88e0.1398t
500 = 88e0.1398t
= e0.1398t
= ln e0.1398t
= 0.1398t
=t
12.4 ≈ t
According to this model, the store will sell 500 phones in
a month in about 12.4 months.
Answer: 12.4 months