462
Exponential and Logarithmic Functions
If nothing else, Example
6.4.1
demonstrates the importance of checking for extraneous solutions
2
when solving equations involving logarithms.
Even though we checked our answers graphically,
extraneous solutions are easy to spot - any supposed solution which causes a negative number
inside a logarithm needs to be discarded. As with the equations in Example
6.3.1
, much can be
learned from checking all of the answers in Example
6.4.1
analytically. We leave this to the reader
and turn our attention to inequalities involving logarithmic functions. Since logarithmic functions
are continuous on their domains, we can use sign diagrams.
Example 6.4.2.
Solve the following inequalities. Check your answer graphically using a calculator.
1.
1
ln(
x
) + 1
≤
1
2. (log
2
(
x
))
2
<
2 log
2
(
x
) + 3
3.
x
log(
x
+ 1)
≥
x
Solution.
1. We start solving
1
ln(
x
)+1
≤
1 by getting 0 on one side of the inequality:
1
ln(
x
)+1
−
1
≤
0.
Getting a common denominator yields
1
ln(
x
)+1
−
ln(
x
)+1
ln(
x
)+1
≤
0 which reduces to
−
ln(
x
)
ln(
x
)+1
≤
0,
or
ln(
x
)
ln(
x
)+1
≥
0.
We define
r
(
x
) =
ln(
x
)
ln(
x
)+1
and set about finding the domain and the zeros
of
r
.
Due to the appearance of the term ln(
x
), we require
x >
0.
In order to keep the
denominator away from zero, we solve ln(
x
) + 1 = 0 so ln(
x
) =
−
1, so
x
=
e
−
1
=
1
e
. Hence,
the domain of
r
is
(
0
,
1
e
)
∪
(
1
e
,
∞
)
. To find the zeros of
r
, we set
r
(
x
) =
ln(
x
)
ln(
x
)+1
= 0 so that
ln(
x
) = 0, and we find
x
=
e
0
= 1. In order to determine test values for
r
without resorting
to the calculator, we need to find numbers between 0,
1
e
, and 1 which have a base of
e
. Since
e
≈
2
.
718
>
1, 0
<
1
e
2
<
1
e
<
1
√
e
<
1
< e
. To determine the sign of
r
(
1
e
2
)
, we use the fact that
ln
(
1
e
2
)
= ln
(
e
−
2
)
=
−
2, and find
r
(
1
e
2
)
=
−
2
−
2+1
= 2, which is (+). The rest of the test values
are determined similarly. From our sign diagram, we find the solution to be
(
0
,
1
e
)
∪
[1
,
∞
).
Graphing
f
(
x
) =
1
ln(
x
)+1
and
g
(
x
) = 1, we see the the graph of
f
is below the graph of
g
on
the solution intervals, and that the graphs intersect at
x
= 1.
0
(+)
1
e
‽
(
−
)
1
0
(+)
y
=
f
(
x
) =
1
ln(
x
)+1
and
y
=
g
(
x
) = 1
2
Recall that an extraneous solution is an answer obtained analytically which does not satisfy the original equation.