The Natural Logarithm

7-29-2008
The Natural Logarithm

Let a be a positive number, a = 1, and let x > 0. The logarithm of x to the base a is the number
y = log
a
x such that a
y
= x. That is,
y = log
a
x means a
y
= x.
Example. What exponential equation is equivalent to log
2
16 = 4?
log
2
16 = 4 is equivalent to 2
4
= 16.
Example. What logarithmic equation is equivalent to 3
4
=
1
81
?
3
4
=
1
81
is equivalent to log
3
1
81
= 4.
Example.
log
2
8 = 3 because 2
3
= 8.
log
5
1
25
= 2 because 5
2
=
1
25
.
log
2
1
16
= 4 because 2
4
=
1
16
.
log
32
2 =
1
5
because 32
1/5
= 2.
Remember that e is the number whose decimal value is 2.718281828459045 . . .. According to the deni-
tion above,
y = log
e
x means e
y
= x.
Logs to the base e are so important that they have a special name; they are called natural logarithms.
They also have a special symbol: Write lnx in place of log
e
x.
On most calculators, lnx and e
x
are on the same button. For example, you can use your calculator to
verify that
ln 2 0.69315 and e
3
20.08554.
The graph of the natural logarithm is shown below:
2 4 6 8 10
-4
-3
-2
-1
1
2
y = ln x
1
lnx and e
x
are inverse functions if you do one, then the other, you get what you started with. In
symbols,
lne
a
= a for all a,
e
lnb
= b for b > 0.
These relationships are often useful for solving equations involving e
x
or lnx.
In addition, lnx satises the usual properties of logarithms.
Properties of the natural logarithm.
(a) ln 1 = 0. (This makes sense, since e
0
= 1.)
(b) ln(ab) = lna + lnb.
(c) ln
a
b
= lna lnb.
(d) lna
p
= p lna.
Example. Solve 4
x
= 5.
To get something out of an exponent, take logs:
ln 4
x
= ln 5, xln4 = ln 5, x =
ln 5
ln 4
1.16096.
Example. Solve lnx + ln(x 5) = ln 6.
To get something out of a log, use e
x
:
lnx(x 5) = ln 6, e
lnx(x5)
= e
ln 6
, x(x 5) = 6, x
2
5x = 6,
x
2
5x 6 = 0, (x 6)(x + 1) = 0, x = 6 or x = 1.
Check the possible solutions by plugging back in:
ln 6 + ln(6 5) = ln 6 + ln 1 = ln 6. (Checks!)
ln(1) + ln(1 5) is undened.
Therefore, the only solution is x = 6.
Example. $1000 is to be invested at 4% annual interest, compounded quarterly. For how many years must
the investment be held to accrue to at least $10000?
Let n be the number of years required. Then
10000 = 1000
1 +
0.04
4
4n
, 10 =
1 +
0.04
4
4n
, ln 10 = ln (1.01)
4n
,
ln 10 = 4nln1.01, n =
ln 10
4 ln1.01
57.85197 years.
2
Example. Write log
3
5
x
6
y
4
z
3
in terms of log
3
x, log
3
y, and log
3
z.
log
3
5
x
6
y
4
z
3
=
1
5
(6 log
3
x + 4 log
3
y 3 log
3
z) .
Example. Write 4 lnx + 3 ln 2 6 lny as a single log.
4 lnx + 3 ln2 6 lny = ln
8x
4
y
6
.
Example. Solve 2
2x
= 3.
To get something out of an exponent (the 2x), take logs:
ln 2
2x
= ln 3, 2xln 2 = ln 3.
Then
2xln2 = ln 3
/ 2 ln 2 2 ln2
x =
ln 3
2 ln2
The solution is x =
ln 3
2 ln 2
0.79248.
Example. Solve 5
2x
= 5
3x2
.
Take logs on both sides:
ln 5
2x
= ln 5
3x2
, 2xln 5 = (3x 2) ln5.
Then
2xln5 = (3x 2) ln5
/ ln 5 ln 5
2x = 3x 2
And so
2x = 3x 2
3x 3x
x = 2
1 1
x = 2
The solution is x = 2.
Example. How many years will it take $5000 invested at 4.8% annual interest, compounded monthly, to
accrue to $10000?
Plug P = 5000, A = 10000, r = 0.048, k = 12 in the compound interest formula
A = P
1 +
r
k
nk
.
3
I get
10000 = 5000
1 +
0.048
12
12n
, 10000 = 5000 1.004
12n
.
Then
10000 = 5000 1.004
12n
/ 5000 5000
2 = 1.004
12n
Take logs on both sides:
ln 2 = ln 1.004
12n
, ln 2 = 12nln1.004.
So
ln 2 = 12nln 1.004
/ 12 ln 1.004 12 ln 1.004
ln 2
12 ln 1.004
= n
Thus, n =
ln 2
12 ln 1.004
14.46943 years.
Example. Solve log
x
1
81
= 4.
log
x
1
81
= 4 means that x
4
=
1
81
. Therefore,
x =
4
1
81
=
1
3
.
The base of a logarithm must be positive, so x =
1
3
.
Example. Compute log
2
3 on your calculator.
Your calculator can compute logs to the base 10 and natural logs, which are logs to the base e. To
compute logs to other bases, use the conversion formula
log
a
b =
log
c
b
log
c
a
.
You t ake c to be a base available on your calculator. For example, using natural logs,
log
a
b =
lnb
lna
.
So in this case,
log
2
3 =
ln 3
ln 2
1.58496.
Example. Solve lnx + ln(x 1) = ln(x + 3).
lnx + ln(x 1) = ln(x + 3), lnx(x 1) = ln(x + 3), e
ln x(x1)
= e
ln(x+3)
,
x(x 1) = x + 3, x
2
x = x + 3.
4
Then
x
2
x = x + 3
x 3 x 3
x
2
2x 3 = 0
Factor and solve:
x
2
2x 3 = 0
(x 3)(x + 1) = 0

x 3 = 0 x + 1 = 0
x = 3 x = 1
Check: If x = 3,
ln x + ln(x 1) = ln 3 + ln 2 = ln 6 = ln(x + 3).
But if x = 1, lnx is undened.
Hence, the only solution is x = 3.
Example. Solve for x: 2 lnx + ln(x 4) = ln 5 + lnx.
2 lnx + ln(x 4) = ln 5 + lnx, lnx
2
+ ln(x 4) = ln 5 + lnx, lnx
2
(x 4) = ln 5x,
e
(ln x
2
(x4))
= e
(ln 5x)
, x
2
(x 4) = 5x, x
3
4x
2
= 5x.
Then
x
3
4x
2
= 5x
5x 5x
x
3
4x
2
5x = 0
Factor and solve:
x
3
4x
2
5x = 0
x(x 5)(x + 1) = 0

x = 0 x 5 = 0 x + 1 = 0
x = 5 x = 1
Check: x = 0 and x = 1 cant be substituted into the original equation, because you cant take the
log of 0 or a negative number.
If x = 5,
2 lnx + ln(x 4) = 2 ln5 + ln(5 4) = 2 ln 5, ln 5 + lnx = ln 5 + ln 5 = 2 ln5.
The only solution is x = 5.
Example. Solve for x: e
2x
7e
x
8 = 0.
Since e
2x
= (e
x
)
2
, the equation is
(e
x
)
2
7e
x
8 = 0.
Factor and solve:
(e
x
)
2
7e
x
8 = 0
(e
x
8)(e
x
+ 1) = 0

e
x
8 = 0 e
x
+ 1 = 0
e
x
= 8 e
x
= 1
5
e
x
= 1 is impossible, because e
x
> 0 for all x.
To solve e
x
= 8, take logs of both sides:
e
x
= 8, lne
x
= ln 8, x = ln 8.
The only solution is x = ln 8.
Example. How much must be invested at 3.6% annual interest, compounded monthly, to accrue to $3000
after 2 years?
Using the compound interest formula,
3000 = P
1 +
0.036
12
212
, 3000 = P
1.003
24
.
Therefore,
P =
3000
1.003
24
2791.89359.
Example. If log
a
x = 1.7 and log
a
y = 2.6, compute log
a
x
3
y
.
log
a
x
3
y
= log
a
x
3
log
a
y = 3 log
a
x
1
2
log
a
y = 3 1.7
1
2
2.6 = 3.8.
Example. Find the domain of f(x) = ln(x 1)(x + 2).
You cant take the log of 0 or a negative number, so f is only dened if (x 1)(x + 2) > 0.
y = (x 1)(x + 2) is a parabola opening upward with roots at x = 1 and x = 2.
-2 1
The domain is x < 2 or x > 1.
Example. Solve for x: (lnx)
2
2 lnx = 35.
(lnx)
2
2 lnx = 35
35 35
(lnx)
2
2 lnx 35 = 0
6
Factor and solve:
(lnx)
2
2 lnx 35 = 0
(lnx 7)(lnx + 5) = 0

lnx 7 = 0 lnx + 5 = 0
lnx = 7 lnx = 5
Solve the two equations by exponentiating:
lnx = 7, e
(ln x)
= e
7
, x = e
7
.
lnx = 5, e
(lnx)
= e
5
, x = e
5
.
The solutions are x = e
7
and x = e
5
.
Example. Solve for x: 4
2x
= 6
x+1
.
4
2x
= 6
x+1
, ln 4
2x
= ln 6
x+1
, 2xln 4 = (x + 1) ln6, (2 ln4)x = (ln 6)x + ln 6.
Then
(2 ln 4)x = (ln6)x + ln 6
(ln 6)x (ln6)x
(2 ln 4)x (ln 6)x = ln 6
Then
(2 ln4)x (ln6)x = ln 6, (2 ln4 ln 6) x = ln 6, x =
ln 6
2 ln 4 ln 6
1.82678.
c 2008 by Bruce Ikenaga 7

The Natural Logarithm

Uploaded by

Document Informationclick to expand document information

Copyright:

Available Formats

The Natural Logarithm

Uploaded by

Document Information

Original Description:

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

The Natural Logarithm

Uploaded by

Copyright:

Available Formats

7-29-2008

The Natural Logarithm

You might also like