Download Section 6.6 – Logarithmic and Exponential Equations

184
Section 6.6 – Logarithmic and Exponential 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)
( MN ) = log (M) – log (N)
b)
loga
c)
loga(Mr) = r loga(M)
a
a
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:
Solve 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.
Solve the following:
Ex. 1
log5(3 – 4x) = 2
185
Solution:
Since log5(3 – 4x) is defined when 3 – 4x > 0 or x < ¾, then the
domain is (– ∞, ¾). Using the Loga Definition, we can rewrite the
equation as an exponential equation:
log5(3 – 4x) = 2
(apply the Loga Definition)
2
3 – 4x = 5
3 – 4x = 25
(solve for x)
– 4x = 22
x = – 5.5
Since – 5.5 is in the domain of the function, the solution is {– 5.5}.
Ex. 2
– 2 log6(x) = log6(16)
Solution:
Since log6(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
2
x =
(use the square root property)
x=
16
1
±
4
1
Since the domain is (0, ∞), our only solution is { }.
4
€
Ex. 3
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
1
(– , ∞) ∩ (– 2, ∞) = (– , ∞).
6
€
6
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)
5
1
x=–
or x =
2
Since the domain is (–
€
3
1
,
6
1
3
∞), our only solution is { }.
€
€
€
186
Ex. 4
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. 5
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 #2:
Solve Exponential Equations.
To solve exponential equations, we try to get our equations in either the
form au = av and apply the ax is 1-to-1 theorem or the form au = M and apply
the loga is 1-to-1 by taking the natural log of both sides.
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Solve the following. Give both the exact and approximate answers
when appropriate:
Ex. 6
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:
2x – 5 = 24
(apply the ax is 1-to-1 thm.)
x–5=4
(solve)
x=9
The solution is {9}.
Ex. 7
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)
ln(2.8)
x=–
≈ – 0.7427
ln(4)
{
The solution is –
Ex. 8
ln(2.8)
ln(4)
} ≈ {– 0.7427}
€ 23x – 4 = 71 + 2x
Solution:
€ is all real numbers, so we have no restrictions.
The domain
23x – 4 = 71 + 2x
(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}.
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Ex. 9
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)
ln(10)
x=
≈ 2.0959
ln(3)
≈ {2.0959}.
{ ln(10)
ln(3) }
The solution is
log2(x) = 2x
Ex. 10a
€
Ex. 10b
log1/2(x) =
1 x
2
( )
Solution:
€
This problem
is extremely difficult to solve algebraically, but very easy
if we remember the graphs of y = loga(x) and y = ax:
€
4
4
3
2
y=2
y=
x
1
x
( 21 )
2
1
y = log2(x)
0
-4
-3
-2
-1
-1
0
0
1
2
3
4
€
-4
-3
-2
-1
-1
-2
-2
-3
-3
-4
y = log1/2(x)
3
x
0
1
2
-4
Thus, log2(x) = 2 has no solution and the solution to log1/2(x) =
is ≈ {2/3}.
€
3
1 x
2
( )
4