2 ln x 1 ln4 and g x 2 ln x ln2 we see the graphs

2 ln( x +1)
ln(4)
and g ( x ) =
2 ln( x )
ln(2)
, we see the
graphs intersect only at x = 1 +
√
3 ≈ 2 . 732. The solution x = 1 −
√
3 < 0, which means if
substituted into the original equation, the term 2 log
2
(
1 −
√
3
)
is undefined.
y = f ( x ) = 1 + 2 log
4
( x + 1) and y = g ( x ) = 2 log
2
( x )

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.

6.4 Logarithmic Equations and Inequalities 463
2. Moving all of the nonzero terms of (log
2
( x ))
2
< 2 log
2
( x ) + 3 to one side of the inequality,
we have (log
2
( x ))
2
− 2 log
2
( x ) − 3 < 0. Defining r ( x ) = (log
2
( x ))
2
− 2 log
2
( x ) − 3, we get
the domain of r is (0 , ∞ ), due to the presence of the logarithm. To find the zeros of r , we
set r ( x ) = (log
2
( x ))
2
− 2 log
2
( x ) − 3 = 0 which results in a ‘quadratic in disguise.’ We set
u = log
2
( x ) so our equation becomes u
2
− 2 u − 3 = 0 which gives us u = − 1 and u = 3. Since
u = log
2
( x ), we get log
2
( x ) = − 1, which gives us x = 2
− 1
=
1
2
, and log
2
( x ) = 3, which yields
x = 2
3
= 8. We use test values which are powers of 2: 0