Putting these two steps together, we simplify
ln
(
e

2
t
)
= ln
(
1
9
)
to

2
t
=

ln(9). We arrive at our solution,
t
=
ln(9)
2
which simplifies to
t
= ln(3).(Can you explain why?)The calculator confirms the graphs off(x) = 75 and
≈
1
.
099.
y
=
f
(
x
) = 9
·
3
x
and
y
=
f
(
x
) = 75 and
y
=
g
(
x
) = 7
2
x
y
=
g
(
x
) =
100
1+3
e

2
x
5. We start solving 25
x
= 5
x
+ 6 by rewriting 25 = 5
2
so that we have
(
5
2
)
x
= 5
x
+ 6, or
5
2
x
= 5
x
+ 6. Even though we have a common base, having two terms on the right hand side
of the equation foils our plan of equating exponents or taking logs. If we stare at this long
enough, we notice that we have three terms with the exponent on one term exactly twice that
of another. To our surprise and delight, we have a ‘quadratic in disguise’. Letting
u
= 5
x
,
we have
u
2
= (5
x
)
2
= 5
2
x
so the equation 5
2
x
= 5
x
+ 6 becomes
u
2
=
u
+ 6. Solving this as
u
2

u

6 = 0 gives
u
=

2 or
u
= 3. Since
u
= 5
x
, we have 5
x
=

2 or 5
x
= 3. Since
5
x
=

2 has no real solution, (Why not?) we focus on 5
x
= 3. Since it isn’t convenient to
express 3 as a power of 5, we take natural logs and get ln (5
x
) = ln(3) so that
x
ln(5) = ln(3)
7.3 Exponential Equations and Inequalities
365
or
x
=
ln(3)
ln(5)
. When we graph
f
(
x
) = 25
x
and
g
(
x
) = 5
x
+ 6, we see that they intersect at
x
=
ln(3)
ln(5)
≈
0
.
6826.
6. At first, it’s unclear how to proceed with
e
x

e

x
2
= 5, besides clearing the denominator to
obtain
e
x

e

x
= 10. Of course, if we rewrite
e

x
=
1
e
x
, we see we have another denominator
lurking in the problem:
e
x

1
e
x
= 10. Clearing this denominator gives us
e
2
x

1 = 10
e
x
,
and once again, we have an equation with three terms where the exponent on one term is
exactly twice that of another  a ‘quadratic in disguise.’ If we let
u
=
e
x
, then
u
2
=
e
2
x
so the
equation
e
2
x

1 = 10
e
x
can be viewed as
u
2

1 = 10
u
. Solving
u
2

10
u

1 = 0, we obtain
by the quadratic formula
u
= 5
±
√
26. From this, we have
e
x
= 5
±
√
26. Since 5

√
26
<
0,
we get no real solution to
e
x
= 5

√
26, but for
e
x
= 5 +
√
26, we take natural logs to obtain
x
= ln
(
5 +
√
26
)
. If we graph
f
(
x
) =
e
x

e

x
2
and
g
(
x
) = 5, we see that the graphs intersect
at
x
= ln
(
5 +
√
26
)
≈
2
.
312
y
=
f
(
x
) = 25
x
and
y
=
f
(
x
) =
e
x

e

x
2
and
y
=
g
(
x
) = 5
x
+ 6
y
=
g
(
x
) = 5
The authors would be remiss not to mention that Example
7.3.1
still holds great educational
value. Much can be learned about logarithms and exponentials by verifying the solutions obtained
in Example
7.3.1
analytically.
For example, to verify our solution to 2000 = 1000
·
3

0
.
1
t
, we
substitute
t
=

10 ln(2)
ln(3)
and obtain
2000
?
=
1000
·
3

0
.
1
“

10 ln(2)
ln(3)
”
2000
?
=
1000
·
3
ln(2)
ln(3)
2000
?
=
1000
·
3
log
3
(2)
Change of Base
2000
?
=
1000
·
2
Inverse Property
2000
X
=
2000
The other solutions can be verified by using a combination of log and inverse properties. Some
fall out quite quickly, while others are more involved. We leave them to the reader.
366
Exponential and Logarithmic Functions
Since exponential functions are continuous on their domains, the Intermediate Value Theorem
4.1
applies. As with the algebraic functions in Section
??
, this allows us to solve inequalities using
sign diagrams as demonstrated below.