Download Section 4.5 – Solving Exponential and Logarithmic Equations

181
Section 4.5 – Solving Exponential and Logarithmic
Equations
In this section, we will use the following tools to help us solve exponential
and logarithmic equations:
Tool 1: Loga Definition
For x > 0 and a > 0 and a ≠ 1, then
y = f(x) = loga(x) if and only if x = ay.
Tool 2: Loga is 1-to-1 Theorem
Let M, N, and a be positive real numbers and a ≠ 1, then
M = N if and only if loga(M) = loga(N).
Tool 3: ax is 1-to-1 Theorem
Let u and v be real numbers and a > 0 and a ≠ 1, then
u = v if and only if au = av
We will also use our properties of logarithms to help solve equations:
Tool 4: Law of Logs Theorem
Let M > 0, N > 0, a > 0, a ≠ 1, and r be any real number. Then
a)
loga(M•N) = loga(M) + loga(N)
b)
c)
loga
( MN ) = log (M) – log (N)
a
a
r
loga(M ) = r loga(M)
We will need to be cautious in using these properties since that even
though €
log of a product and/or quotient may be defined, the log of each
factor may not be, so we will need to check our answers with the original
equation. Also, using loga(Mr) = r loga(M) when r is an even integer can
leave out possible solutions or introduce extraneous ones.
Objective 1:
Solving Exponential Equations
Solve the following. Give both the exact and approximate answers
when appropriate:
Ex. 1
3•2x – 5 = 48
Solution:
The domain is all real numbers, so we have no restrictions.
First, divide both sides by 3 to get 2x – 5 = 16.
Since 16 = 24, rewrite the equation as:
182
2x – 5 = 24
x–5=4
x=9
The solution is {9}.
Ex. 2
4 – x – 2.8 = 0
Solution:
The domain is all real numbers, so we have no restrictions.
4 – x – 2.8 = 0
(add 2.8 to both sides)
–x
4 = 2.8
(use Loga is 1-to-1 thm by taking ln of both sides)
–x
ln(4 ) = ln(2.8)
(apply the Law of Logs, part c)
– x ln(4) = ln(2.8)
(solve for x)
x=–
ln(2.8)
ln(4)
{
The solution is –
Ex. 3
(apply the ax is 1-to-1 thm.)
(solve)
≈ – 0.7427
ln(2.8)
ln(4)
} ≈ {– 0.7427}
€ 23x – 4 = 71 + 2x
Solution:
€ is all real numbers, so we have no restrictions.
The domain
3x – 4
1 + 2x
2
=7
(use Loga is 1-to-1 thm by taking ln of both sides)
3x – 4
1 + 2x
ln(2
) = ln(7
)
(apply the Law of Logs, part c)
(3x – 4) ln(2) = (1 + 2x) ln(7)
(distribute)
3x ln(2) – 4 ln(2) = ln(7) + 2x ln(7)
(add 4 ln(2) – 2x ln(7) to both sides)
3x ln(2) – 2x ln(7) = ln(7) + 4 ln(2)
(factor x out)
[3 ln(2) – 2 ln(7)]x = ln(7) + 4 ln(2)
(divide both sides by 3 ln(2) – 2 ln(7))
ln(7)+4ln(2)
x=
≈ – 2.6035
The
3ln(2)−2ln(7)
ln(7)+4ln(2)
solution is
3ln(2)−2ln(7)
{
} ≈ {– 2.6035}.
Ex. 4 €
e2x – 9ex + 20 = 0
Solution:
€
The domain
is all real numbers, so we have no restrictions.
This equation almost appears to be quadratic since e2x = (ex)2
e2x – 9ex + 20 = 0
(rewrite e2x as (ex)2)
(ex)2 – 9ex + 20 = 0
(let w = ex)
w2 – 9w + 20 = 0
(factor)
183
(w – 4)(w – 5) = 0
(solve)
w = 4 or w = 5
(replace w by ex)
ex = 4 or ex = 5
(rewrite as logarithmic equations)
x = ln(4) or x = ln(5)
The solution is {ln(4), ln(5)} ≈ {1.3863, 1.6094}
Ex. 5
32x – 3x + 2 – 10 = 0
Solution:
The domain is all real numbers, so we have no restrictions.
This equation almost appears to be quadratic since 32x = (3x)2 and
3x + 2 = 3x•32 or 9•3x. Let's try letting w = 3x. Thus, we get:
32x – 3x + 2 – 10 = 0
(3x)2 – 9•3x – 10 = 0
(replace 3x by w)
w2 – 9w – 10 = 0
(factor)
(w – 10)(w + 1) = 0
(solve for w)
w = 10 or w = – 1
(replace w by 3x)
3x = 10 or 3x = – 1
(take the ln of both sides)
x
x
ln(3 ) = ln(10) or ln(3 ) = ln(– 1)
Since ln(– 1) is undefined, the second equation has no solution.
ln(3x) = ln(10)
(apply the Laws of Logs, part c)
x ln(3) = ln(10)
(solve for x)
x=
ln(10)
ln(3)
Objective 2:
≈ 2.0959
The solution is
≈ {2.0959}.
{ ln(10)
ln(3) }
Finding the Domain of Logarithmic Functions.
Since the
€ logarithmic function is the inverse
€ function of the exponential
function, the domain of a logarithmic function is equal to the range of the
exponential function, which is (0, ∞). In other words, for a logarithmic
function to be defined, the argument has to be greater than zero.
Find the domain of the following:
Ex. 6a
f(x) = log3(2 – 3x)
Ex. 6b
g(x) = loge(11x + 30.25)
Ex. 6c
Ex. 6d
h(x) = log4(4 – x2)
g(x) = log0.5
( x−3
x+2 )
Solution:
a)
The argument has to be greater than zero:
2 – 3x > 0 (solve)
€
2
2
x<
The domain is (– ∞, ).
3
€
3
€
184
b)
The argument has to be greater than zero.
11x + 30.25 > 0 (solve)
x > – 2.75
The domain is (– 2.75, ∞).
c)
The argument has to be greater than zero.
x−3
x+2
€
>0
3 is a zero.
x = – 2 is a V.A.
We will need a sign chart:
Interval:
Value:
–2
(– ∞, – 2)
–3
3
(– 2, 3)
0
(3, ∞)
4
f(x):
f(– 3) = 6
f(0) = – 1.5
f(4) =
Sign:
+
–
+
1
6
Thus, the domain is (– ∞, – 2) U (3, ∞)
€
d)
The argument has to be greater than zero.
4 – x2 > 0
(2 – x)(2 + x) > 0
– 2 and 2 are zeros
We will need a sign chart:
Interval:
Value:
f(x):
Sign:
–2
(– ∞, – 2)
–3
f(– 3) = – 5
–
2
(– 2, 2)
0
f(0) = 4
+
(2, ∞)
3
f(3) = – 5
–
Thus, the domain is (– 2, 2).
Objective 3:
Solving Logarithmic Equations.
To solve logarithmic equations, we need to get a our equation into either
the form of loga(M) = loga(N) and apply the Loga is 1-to-1 theorem or the
form y = loga(x) and apply the Loga definition.
185
Solve the following:
Ex. 7
log5(3 – 4x) = 2
Solution:
Since log5(3 – 4x) is defined when 3 – 4x > 0 or x < ¾, then the
domain is (– ∞, ¾).
log5(3 – 4x) = 2
(apply the Loga Definition)
2
3 – 4x = 5
3 – 4x = 25
(solve for x)
x = – 5.5
Since – 5.5 is in the domain of the function, the solution is {– 5.5}.
Ex. 8
– 2 log6(x) = log6(16)
Solution:
Since log5(x) is defined when x > 0, then the domain is (0, ∞).
We can rewrite – 2 log6(x) as log6(x – 2) using The Law of Logs, part c.
– 2 log6(x) = log6(16) (apply The Law of Logs, part c)
log6(x – 2) = log6(16)
(apply the Loga is 1-to-1 Thm)
–2
x = 16
(take the "reciprocal of both sides")
1
16
1
±
4
x2 =
x=
(use the square root property)
1
Since the domain is (0, ∞), our only solution is { }.
4
€
Ex. 9
log49(6x + 1) + log49(x + 2) = ½
€
Solution:
Since log49(6x + 1) is defined when 6x €
+ 1 > 0 or x > –
1
6
and since
log49(x + 2) is defined when x + 2 > 0 or x > – 2, the domain is
(–
1
,
6
€
∞) ∩ (– 2, ∞) = (–
∞).
log49(6x + 1) + log49(x + 2) = ½ (apply €
the Law of Logs, part a)
log49[(6x + 1)(x + 2)] = ½
(FOIL)
2
log49[6x + 13x + 2] = ½
(apply the Log Definition)
2
1/2
€
6x + 13x + 2 = 49
(simplify)
2
6x + 13x + 2 = 7
(subtract 7 from both sides)
2
6x + 13x – 5 = 0
(factor)
(2x + 5)(3x – 1) = 0
(solve)
x=–
5
2
or x =
Since the domain is (–
€
1
,
6
1
3
1
,
6
1
3
∞), our only solution is { }.
€
€
€
186
Ex. 10
ln(3x) – ln(x – 8) = ln(x + 5)
Solution:
Since ln(3x) is defined when x > 0, ln(x – 8) is defined when x – 8 > 0
or when x > 8, and ln(x + 5) is defined when x + 5 > 0 or x > – 5, the
domain is (0, ∞) ∩ (8, ∞) ∩ (– 5, ∞) = (8, ∞).
ln(3x) – ln(x – 8) = ln(x + 5) (apply the Law of Logs, part b)
ln
3x
( x−8
) = ln(x + 5)
3x
x−8
=x+5
(apply the Loga is 1-to-1 Thm)
(multiply both sides by x – 8)
3x = (x + 5)(x – 8)
(FOIL)
2
€
3x = x – 3x – 40
(subtract 3x from both sides)
2
0 = x – 6x – 40
(factor)
€
0 = (x – 10)(x + 4)
(solve)
x = 10 or x = – 4
But, x = – 4 is not in the domain, so the solution is {10}.
Ex. 11
log4(x2) = 2
Solution:
Since the log4(x2) is defined when x2 > 0 or when x ≠ 0, the domain is
(– ∞, 0) U (0, ∞).
log4(x2) = 2
(apply the Log Definition)
2
2
x =4
(simplify)
2
x = 16
(use the square root property)
x=±4
Both values are in the domain, so the solution is {– 4, 4}.
Suppose we had use the Law of Logs, part c as our first step. We would
have arrived at the following solution:
log4(x2) = 2
2 log4(x) = 2
(divide both sides by 2)
log4(x) = 1
(apply the Log Definition)
1
x = 4 = 4 (missing an answer)
Hence, notice that does not gives us all of the answers. That is why it is
important to determine the domain before you begin and use great caution
when using the Law of Logs, part c.
Objective 4:
Applications
Recall the formulas for compound Interest:
(
A=P 1+
€
r
n
nt
)
and A = Pert
187
Solve the following:
Ex. 12
How long will it take an investment to triple if it is earns 4%
annual interest:
a)
compounded annually?
b)
continuously?
Solution:
(
a)
r
n
A=P 1+
(
3P = P 1 +
0.04
1
nt
)
1t
)
3 = (1.04)t
t
€
ln(3) = ln[(1.04)
]
ln(3) = t ln(1.04)
€ ln(3)
t=
= 28.011…
(plug in A = 3P, n = 1, and r = 0.04)
(divide both sides by P)
(take the ln of both sides)
(use law of the logs part c)
(solve for t)
ln(1.04)
It will take approximately 28 years for it to triple.
b)
A = Pert
3P = Pe0.04t
3 = e0.04t
0.04t = ln(3)
€
t=
ln(3)
0.04
(plug in A = 3P, and r = 0.04)
(divide both sides by P)
(rewrite as a log equation)
(solve for t)
= 27.4653…
It will take approximately 27.47 years for it to triple.
Ex. 13
The temperature T in Fahrenheit of the engine m minutes after
€
T − 70
ln(
)
200
it was turned off is given by the formula m =
. Find the
−0.08
temperature to the nearest degree of the engine 15 minutes after the
engine was turned off.
Solution:
€
Replace m by 15 and solve:
15 =
ln(
T − 70
200
€
€
(multiply both sides by – 0.08)
−0.08
=e
(
T − 70
200
– 1.2
– 1.2 = ln
T − 70
200
)
– 1.2
)
(rewrite as an exponential equation)
(multiply by 200 & then add 70 to both sides)
T = 200e
+ 70 = 130.238… ≈ 130˚ F
The
€ engine was ≈ 130˚ F after 15 minutes.