Download LOGARITHMS

LOGARITHMS
Definition of Logarithm:
Given:
and
a is a (real) number > 0 but a ≠ 1,
called the
base
of the logarithm,
x is a (real) number > 0.
log a x is defined to be the number y such that:
y
y is the exponent for which a = x .
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_
_
_
_ _
_ _ _ _ _ _ _ _
log a x = y ( log base number = exponent ) may also be read as follows:
y is the logarithm to the base a of x
OR, read another way: x is the antilogarithm to the base a of y.
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_ _ _ _ _ _ _ _ _ _ _
Examples of Evaluating Logarithms by Using the Definition:
Example I: Find log 10 10000.
Solution:
log 10 10000 is the number/exponent y such that
y
10 = 10,000.
In order to find the answer, we start with the number 10 and multiply
it by itself as follows:
if y = 2, then
if y = 3, then
if y = 4, then
2
10 = 10·10 = 100
3
10 = 10·10·10 = 1000
4
10 = 10·10·10·10 = 10,000.
4
Therefore, 10 = 10,000 and we deduce that log 10 10000 = 4.
Answer: log 10 10000 = 4.
We read the answer as follows:
4 is the logarithm to the base 10 of the number 10000.
Example II: Find log 5 3125.
Solution:
log 5 3125 = y is the exponent such that
y
5 = 3125.
After doing a little bit of calculation similar to Example I, we find
5
that 5 = 3125, and, therefore, log 5 3125 = 5.
Answer: log 5 3125 = 5.
We read the answer as follows:
5 is the logarithm to the base 5 of the number 3125.
Example III: Find log 2 (1/4).
Solution:
log 2 (1/4) = y is the exponent such that
y
2 = 1/4.
Proceeding as in the previous examples, we obtain:
1
if y = 1, then
2 =2
2
2 = 2· 2 = 4
if y = 2, then
3
if y = 3, then
2 = 2· 2· 2 = 8
4
if y = 4, then
2 = 2· 2· 2· 2 = 16.
Obviously, by choosing positive exponents, the numbers we obtained above were
all greater than 1/4.
Let’s try 0 and some negative numbers as the exponent y instead.
0
0
If y = 0, then
2 = 1. (From the Laws of Exponents, a = 1 for any base a.)
(-1)
1 = 1/2 .
1
2
(-2)
If y = -2, then
2
= _1 = 1/4 .
2
2
We have now found our answer. Therefore, log
If y = -1, then
2
=
2
(1/4) = -2.
Answer: log 2 (1/4) = -2.
Example IV: Find log 253 1.
Solution:
y
log 253 1 = y is the exponent y such that 253 = 1.
0
(From the Laws of Exponents, we know that a = 1 for any base a .)
Therefore, log 253 1 = 0.
Answer: log 253 1 = 0.
Example V: Find log -7 49.
Solution:
If we proceed in a manner similar to the previous examples, we will
2
conclude that log -7 49 = 2 because (-7) = 49.
However, this answer is FALSE
– not because the arithmetic is incorrect, but because the logarithm
function is NOT DEFINED for bases a ≤ 0.
Answer: log -7 49 is undefined.
2
Example VI: Find log 2 6.
Solution:
We want to find the exponent y such that
2
From Example III, we know that 2 = 4
y
2 = 6.
3
and 2 = 8. We can now surmise that
the exponent y we are looking for is between 2 and 3, but, without a
scientific calculator or a set of log tables, we will not be able to obtain
the answer. Therefore, use your calculator to get the answer.
Answer: log 2 6 = 2.5849625007.
Examples of Evaluating Antilogarithms by Using the Definition:
Example VII: Find the antilogarithm of 3 to the base 10.
Solution:
To find the antilog of 3 means that we are given y = 3 as the exponent
and a = 10 as the base of the logarithm, and we are asked to find the
y
number x such that log a x = y meaning that a = x
3
OR such that log 10 x = 3 meaning that 10 = x.
3
Therefore, x = 10 = 1000 and we deduce that the antilogarithm of 3 to the
base 10 is 1000.
Answer: The antilog of 3 to the base 10 is 1000.
Example VIII: Find the antilogarithm of 7 to the base 2.
Solution:
Without going through the detail, we can
7
antilogarithm of 7 to the base 2 is x = 2 = 128.
simply
say
that
the
Answer: The antilog of 7 to the base 2 is 128.
From the above discussion, we observe that there seems to be an
intimate connection between exponents and logarithms. In fact, there
is a one-to-one correlation between the Laws of Exponents and the
Properties of Logarithms.
Laws of Exponents
a
n
· am
n
a
m
a
(a
n
r
)
0
a
1
a
(n + m)
Properties of Logarithms
1.
log a (N·M ) = log a N + log a M
2.
log a N
M
3.
log a (N )
= r ·log a N
= 1
4.
log a 1
= 0
= a
5.
log a a
= 1
= a
= a
(n - m)
= a
n·r
3
= log a N - log a M
r
We can prove all of the above properties of logarithms by applying the
definition of logs and then using the appropriate Law of Exponents.
For example, Property 5 of logarithms follows from the fifth Law of
Exponents by simply applying the definition of logs.
The first three properties of logarithms are similarly, but not as
easily, obtained. Below is a proof for the second property.
Proposition:
Assuming that a is a (real) number > 0 but a ≠ 1 and N,M are positive
real numbers, prove that:
log a N
M
= log a N - log a M
using the following Law of Exponents:
n
(n - m)
a
= a
(*)
m
a
Proof:
Because we have to prove an equation expressed in full generality, we
begin by letting new variables represent the various pieces of the
equation. In the end, we will piece them together again by trying to
show that the left hand side (LHS) of the equation is equal to the
right hand side (RHS) of the equation.
Let y = log a N and z = log a M.
z
y
Then a = N and a = M by the definition of logs.
(**)
Let w = log a N .
M
w
Then a = N
using the definition of logs
M
y
= a
using (**)
z
a
(y-z)
= a
using the given Law of Exponents (*).
From the fact that the exponential function is one-to-one,
(y-z)
w
if a = a
where the bases are equal, then the exponents must be
equal.
Therefore: w = y – z OR, by substituting the original values for the
w, y, and z, we get:
LHS = log a N = log a N - log a M = RHS.
M
4
q.e.d.
Using the Properties of Logarithms to Solve Problems:
Some hints as to how to recognize when to use the various properties
of the logarithm.
Use
Use
Use
Use
Use
Property
Property
Property
Property
Property
1
2
3
4
5
if:
if:
if:
if:
if:
number
number
number
number
number
is
is
is
is
is
a product of 2 factors
quotient of 2 numbers
a power
equal to 1
equal to the base a
Be aware that if the number is a combination of a product, a quotient,
and a power, then various properties may need to be used in an
appropriate sequence.
ONLINE Logarithm Calculator:
The following web site contains a Logarithm Calculator which will
evaluate both logs and antilogarithms to any positive base a ≠ 1.
http://www.1728.com/logrithm.htm
Example 1: Simplify and evaluate log 80.
Solution:
If the base is not specified, it is assumed to be 10, which is the
common base for the logarithm function. Therefore, the question
becomes:
Simplify and evaluate log 10 80.
Step 1: Simplify by trying to break down the size of the number 80 in
some way and then use one of the properties of logarithms or the
definition of logarithm to simplify further. We first note that 80 can
be factored, keeping in mind that the base is 10.
log 10 80 = log 10 8·10
= log 10 8 + log 10 10
= log 10 8 + 1
(by Property 1 as number is a product)
(by Property 5)
Step 2: Evaluate. Since the logarithm is to the base 10, use a
calculator to evaluate. Therefore:
log 10 80 = log 10 8 + 1 =
0.90308998699 + 1 = 1.90308998699.
Answer: log 10 80 = 1.90309 (rounded to 5 decimal places)
5
Example 2: Simplify and evaluate log 3 (17/6).
Solution:
HINT: Simplify by using property 2 (number is a quotient), followed by
the change of base formula to base 10 (see page 75). Then evaluate
using a calculator.
log 10 (17/6) = log 10 17
6
= log 10 17 - log 10 6
(by Property 2 as number 17/6 was a quotient)
= 1.2304489214 - 0.77815125038
= 0.45229767102
(using your calculator)
Answer: log 10 (17/6) = 0.45230 (rounded to 5 decimal places)
Example 3: Evaluate the number 2
25
using logs.
Solution:
25
Let x = 2 . Take logarithms of both sides of the equation. If the base is
not specified in the problem, assume that the base is 10.
(Actually you may use any base to solve this problem. The only glitch is that you have
to have a way of calculating the logs and antilogarithms to the base that you choose.)
25
Then log 10 x = log 10 2 .
Because the number x is a power, we use Property 3 to obtain:
25
log x = log 10 2 = 25·log 10 2.
Evaluating log 10 2 by using your calculator, we deduce that the exponent
to which the base 10 must be raised to get the number 2 is
0.30102999566. Therefore,
25
log x = log 10 2 = 25·log 10 2 = 25·(0.30102999566) = 7.5257498915.
Because we originally wanted the value of x and we didn’t want to
multiply 2 by itself 25 times, we use the above equation containing
the log of x and solve for x by taking the antilog to the base is 10
of both sides.
Thus:
x = antilog 10(log 10 x)
= antilog 10(7.5257498915)
= 3.3554431992e+7
= 33,554,431.992
= 33,554,432
Answer: 2
25
=
(using your calculator)
(answer in scientific notation)
(by changing to regular notation)
(rounded to the nearest whole number)
33,554,432.
Now, you may argue that the previous question could have done more
easily and faster by multiplying 2 by itself 25 times. However, not
all problems are more easily solved as a straight forward calculation.
6
Example 4: The formula for compound interest is the following:
n
S = P·(1+i)
where P = original investment, i = rate of interest,
n = # of compounding periods, and S = value of investment at maturity.
Suppose you want to invest $10000 and have $15000 after 3 years.
Suppose also that your banker offers you semi-annual compounding on
your investment. What annual interest rate would be necessary in order
to fulfill your request? Is this feasible in today’s financial market?
Solution:
In this problem, P = 10000, S = 15000,
n = 6 (each year has 2 compounding periods since i is an annual interest rate)
and you want to find i.
Putting these values into the formula, we get:
15000 = 10000·(1+i)
and we need to solve for i.
6
First, let us do some preliminary simplification. Recall that if we
perform an operation on one side of the equation, we must perform the
identical operation on the other side so that the equation is still
valid.
Therefore, by dividing both sides of the equation by 10000 and
interchanging sides (so that the unknown is on the LHS), get:
6
(1+i) = 3/2.
th
We could take the 6 root of both sides to eliminate the power 6 on
th
the LHS, but then we would have to have a way of finding the 6 root
of 3/2, which is not trivial or easily accessible. Therefore, let us
resort to using logs.
Taking the log of both sides, we get:
6
log 10(1+i) = log 10(3/2).
By using Property 3 on the LHS and Property 2 on the RHS, we obtain:
6· log 10(1+i) = log 10(3)- log 10(2)
6· log 10(1+i) = 0.47712125472 - 0.30102999566
6· log 10(1+i) = 0.17609125906.
(by evaluating the logs)
To get the factor containing i alone on the LHS, we divide both sides by 6 to get:
log 10(1+i)
= 0.0293485431767.
Now we take antilogs of both sides.
antilog 10(log 10 (1+i)) = antilog 10(0.0293485431767)
(1+i) = 1.0699131939
i
= .0699131939
i
= .0699131939 ·100%
OR
i
(subtract 1 from both sides)
(change decimal to a percentage)
= 7% per annum
Answer: The annual interest rate necessary is 7% per annum.
7
Now let us examine how the application of logs may be used to solve
certain kinds of equations which were not solvable using previously
learned techniques.
x
Example 5: Solve the following equation for x: 2 = 32.
Solution:
Because the unknown x, for which we are solving, is in the exponent
position (i.e. above the fraction line), we need a mechanism to bring
the x down to the fraction line to solve for it. Property 3 of
logarithms accomplishes this.
Therefore, to solve for x, we proceed as follows:
Step 1: Introduce logarithms into the equation by taking logs of both
sides of the equation. Because the base of the exponent in the
original equation is 2, we take log 2 of both sides.
x
log 2 (2 ) = log 2 (32)
Step 2: We try to simplify each side separately.
LHS: use the third property of logs OR apply the definition to
evaluate the LHS.
x
LHS = log 2 (2 )= x · log 2 2 by Property 3 (number contains a power)
= x · 1
by Property 5 (number equals the base a )
NOTE THAT: property 5 is simply an application of the definition of the logarithm.
RHS: apply the definition to evaluate.
RHS = log 2 (32)= y is the exponent such that
RHS = log 2 (32)= 5 .
y
2 = 32. Therefore,
Step 3: Piecing this together, we get the following.
x
LHS = log 2 (2 ) = log 2 (32)= RHS
x
=
5
We have just solved for x.
Answer: The solution to the equation is: x = 5.
8