LOGARITHMS
Objectives:
1.1 There is a particular number, denoted e, which is used extensively in science and mathematics. Its value is approximately 2.718 and may be found as the "e" button on most calculators. Computers generally expect you to enter the number e as Exp(1). Therefore, if you wish to enter the number e^4 on a computer, you would type Exp(4). In the computer algebra system Mathematica, you would type Exp[4]. Note the capital letter and the square brackets around the 4.
1.2 To enter mathematical operations of multiplication and exponentiation on a computer, use the * for multiplication and ^ for exponentiation. Therefore,
3*4 = 12 and 3^4 = 81.
1.3 You may wish to think of the exponentials a^x and e^x as functions of x. You may enter e^x in Mathematica as Exp[x].
The rules for Exp[x] are then expressed as
Exp[x+y] = Exp[x] *Exp[y]
Exp[x]^y = Exp[xy]
Exp[x-y] = Exp[x]/Exp[y]
Exp[0] = 1
Recall 2: If a is different from 0, then a^x is different from 0 for all x.
Recall 3: When two functions are connected by the properties y = f(x) and x = g(y), we say that f(x) is the inverse function of g(y) and g(y) is the inverse function of f(x). Of course this also means that g(f(x)) = x and f(g(y)) = y.
Recall 4: If you plot the graphs of a function and its inverse you will find that they are mirror images of each other when the imaginary mirror is placed along the diagonal line y = x. See illustration below.
The
illustration below shows the functions Exp[x] and
its
inverse.
Introduction 1 (Logarithms):
The inverse function of a^x is called the logarithm
to the
base a and is denoted The graphs of e^x and Ln(x) are displayed
in
the illustration
above. To enter a logarithm to the base a in
Mathematica type Log[a,x].
If you type Log[x], Mathematica will
interpret the base as e. Most
calculators have buttons labeled "log"
and "ln" for Log10(x) and Ln(x),
respectively.
Recall 5 As you would expect, the
rules for manipulating
logarithms look very much like the reverse of those
for manipulating exponents.
They
are:
Let a be a positive number
such that a does
not equal 1, let n be a real number, and
let u and v be positive
real numbers.
Important
Note: The domain of the logarithm
function is the set of all
positive numbers. Therefore, Log(x)
makes no sense when x is
a
negative number.
Recall
6 Since the logarithm and exponential
functions are inverses of
each other, we have the following relationships:
Recall 7 One of the many great
uses of the
logarithm function is to reduce exponents to products.
So, for example,
if you are ever faced with an equation like 4 = 2^(x-3)
and you need to
know what x is, all you have to do is take the logarithm
to the base 2
of both sides of the equation to get
Log2(4) =
Log2[2^(x-3)]
and you
can
proceed by knowing that the left side is equal
to 2 and the right side is
equal to x - 3. Then all you have to do
is solve the equation 2 = x
- 3 to get x = 5.
Recall
8 You know that a^b = a^c if and only if
b = c. Therefore you
may solve equations such as
because you would know that
x^2 + 6 = -x, and that would
mean
x^2 + x + 6 =
0, or
(x + 3)(x + 2) = 0.
This means that x = -3 or x = -2.
Recall 9 You know that Logab
= Logac if and only if b = c.
Therefore you may solve equations of
the form
Log2(x^2 - 6)
= Log2(x) by
what we just said
in the last paragraph.
From
Recall 8, you know that Recall
10 If you know the logarithm to the base
"a" of a number then you know
it to any other base by using the conversion
formula
below:
Calculators often only have
two kinds of log buttons,
log and ln. If you wish to find
Log2(x) then
all you have to do is
find the quotient ln(x)/ln(2). If you are
simply converting from
common logarithms (base 10) to natural logarithms
(base e) or back, you
could use the conversion factors given
by
Ln(x) = 2.30259 Lg(x), or Lg(x) = .43429
ln(x).
Example
1. Example 2. Example
3. The expression
can be written as a single logarithm
Example
4. To calculate
Log5(1/3) you may have
to calculate the
ratio -Ln(3)/Ln(5)
Example
5. To solve
an equation Log2(x - 4) = 1 -
Log2(x - 3) you would collect
the logarithms
to
one side to get Log2(x
- 4) +
Log2(x - 3) = 1, then
group the
logarithms together to get Log2((x
- 4)(x - 3)) = 1. Now recall that if Loga(y)
= 1, then y = a. Therefore (x -
4)(x-3)=2, which gives x = 2 or x = 5. But x = 2 cannot
be a
solution because Log2(x - 4) is the
logarithm of a negative number
when x = 2 and there is no such thing as
the logarithm of a negative number.
So the only solution is x =
5.
Example
6. What if
you want to find x when
Log2(x
- 4) + Log2(x - 3) = 2, then
collect the logarithms to
get Log2((x
- 4)(x - 3)) =
2. Now use the fact that a^b = a^c if and only if
b = c to see
that
Therefore, (x - 4)(x - 3) = 4. If
you multiply
out the left side you get x^2 - 7x - 12 = 4. The
equation then becomes
x^2 - 7x - 16 = 0. This equation does not
factor; so we have to use
the quadratic formula to get the two
solutions.
But one of the solutions is less than 3
and therefore is
not in the domain
of Log2(x - 3). So we
have only one
valid solution.
Note: log(x) is
a shorthand notation for the logarithm to the base 10 of x. 2. If
Log4(x)
= 5/2, what is
x?
3 Simplify
Log5(5)
4 Simplify Exp[ln7].
5 Write
Log3(x
+ 2) + 2Log3x - Log32
as a single logarithm
6 Use natural
logarithms to evaluate Log527
7 If Loga(8/27)
= 3, what is a?
8 Solve the equation 9^x = 12 for x.
9
Solve the equation 7^(2x-1) = 7^(x+2) for x.
10.
Solve the equation 4^(2x-1) = 3^(x+2) for x.
11
Solve the equation exp(x^2)=100
for
x.
12 Solve the equation 5 =
2(3 - e^x) for x.
13 Solve
the equation Logx(x
+ 6) -
Logx(x + 2) = Logxx
for x.
14 Solve
the equation Log3(3x^2)^2
- 1 = 3 for x.
15 If y =
Log2(x^2
+ 6), write x as a
function of y.
16 Solve the equation (Logx)^2 =
Log(x^2) for x.
1. 6
2. x = 32
3. 1
4. 7
5. Log3[x^2(x+2)/2]
6.
Ln(27)/Ln(5)
7. a = 2/3
8. Log912
9. x = 3
10.. x
= 2 Log43
+ 1
11. x = ±sqrt(Ln(100)
12. x =
-Ln(2)
13. x = 2
14. x =
sqrt(3)
15. x = ± sqrt(2)^y -
6
16. x = 1 or x = 100
The inverse
function of e^x
is called the natural
logarithm of x and is often denoted
Ln(x).
Rules of
Logarithms
From this it follows that
x^2 - 6 = x and therefore x
= 3 or x = -2. But x cannot equal -2
because when we check it we
find that we have to make sense of Log
(-2). The logarithm function
does not have negative numbers in its
domain. (See the important
note above.) So the only
solution is x = 3.
Examples
EXERCISES
1.
Find 
ANSWERS