Rules for logarithms
We review the properties of logarithms from the previous lecture. In that
lecture, we developed the following identities.
The “Product Property” is an identity involving the logarithm of product:
Elementary Functions
logb (M N ) = logb M + logb N
Part 3, Exponential Functions & Logarithms
Lecture 3.4a, Working With Logarithms
(1)
The “Quotient Property” is an identity involving the log of quotient:
Dr. Ken W. Smith
logb ( M
N ) = logb M − logb N
(2)
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The “Exponent Property” allows us to rewrite the log of an expression
with an exponent:
logb (M c ) = c · logb M
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Exponential Functions
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Rules for logarithms
There is the change of base equation: if c is a positive real number then
logb x =
logc x
logc b
(4)
The first three equations here are properties of exponents translated into
“logarithm language.”
The fourth equation allows us to choose the base of our logarithm.
There are two identities that express the inverse relationship between
exponential and logarithmic functions:
logb
(bx )
= x.
(5)
blogb x = x.
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The last two equations in the list identify the logarithm as the inverse
function of the exponential function.
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Practicing Logarithms
Practicing Logarithms
Let’s practice these properties of logarithms on some exercises.
2
Some worked problems. Expand the following expressions.
e3
1 ln 2
x +4
By the quotient property
ln
(5(x+1))3
2 ln
.
(5x−7)2
3
2
100x y
3 log
.
z4
3
2
5
4 log(x y z )12 .
(5(x + 1))3
= ln((5(x + 1))3 ) − ln((5x − 7)2 ).
(5x − 7)2
By the exponent property
ln((5(x + 1))3 ) − ln((5x − 7)2 ) = 3 ln(5(x + 1)) − 2 ln(5x − 7).
Solutions.
1 By the quotient property
ln(
e3
) = ln e3 − ln(x2 + 4).
x2 + 4
By the product property this is equal to
3(ln 5+ln(x+1))−2 ln(5x−7) = 3 ln 5 + 3 ln(x + 1) − 2 ln(5x − 7) .
By the first inverse property, since ln() stands for the logarithm base
e, then ln e3 = 3 so the answer is
3 − ln(x2 + 4) .
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Practicing Logarithms
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Practicing Logarithms
By the quotient property,
4
100x3 y 2
= log 100x3 y 2 − log z 4 .
log
z4
By the exponent property,
log(x3 y 2 z 5 )12 = 12 log(x3 y 2 z 5 )
By the product property
By the product property,
3 2
3
2
log 100x y = log 100 + log x + log y .
12 log(x3 y 2 z 5 ) = 12(log x3 + log y 2 + log z 5 )
By the exponent property, we can rewrite all the exponents so that
log 100 + log x3 + log y 2 − log z 4 = log 100 + 3 log x + 2 log y − 4 log z.
and then by the exponent property we have
12(3 log x + 2 log y + 5 log z) .
Since log() stands for the logarithm base ten then log 100 = 2 and so
our final answer is
(Also acceptable is 36 log x + 24 log y + 60 log z .)
2 + 3 log x + 2 log y − 4 log z .
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Simplification of logarithms
Simplification of logarithms
We can use our six logarithm identities to simplify expressions involving
logs. Here are some worked examples.
Use properties of logarithms to simplify the following expressions.
√
More worked problems.
Use properties of logarithms to simplify the following expressions.
3
1
logb ( bb5 )
2
logb ((b3 )5 )
2
loge (e
4
10log10 (5)
5
e− ln 3 ,
Solutions.
Solutions.
1
2)
3
b3
b5
We could simplify
= b−2 and then recognize that logb (b−2 ) = −2.
Or we could use the quotient property of logs and the first inverse
3
property to compute logb ( bb5 ) = logb b3 − logb b5 = 3 − 5 = −2.
By the exponent property logb ((b3 )5 ) = 5 logb b3 . By the first inverse
property 5 logb b3 = 5(3) = 15.
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Simplification of logarithms
3
4
5
√
By the first inverse property loge (e
By the second inverse property,
2)
10log10 (5)
By the exponent property e− ln 3 = e
1
inverse property, eln 3 = 13 .
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=
√
= 5.
ln(3−1 )
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2.
1
= eln( 3 ) . By the first
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Logarithms
Sometimes a problem has an answer in a base which is intrinsic to the
problem but it is not a base with which we can easily do computations. In
that case we need to be prepared to change the base to one with which we
can compute.
Worked problems on changing the base of the logarithm.
Use the “change of base” identity to write the following as fractions
involving ln(). Use a calculator (or computer software program) to
approximate the answer.
1 log2 (5).
2 log2 (125).
3 log16 (17).
4 log(5).
5 log2 (1024).
Solutions.
ln 5
1 log2 (5) = ln 2 ≈ 2.3219 .
ln 125
2 log2 (125) = ln 2 ≈ 6.9658 .
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ln 17
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In the next presentation we continue to practice our logarithm properties.
(END)
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Simplification of logarithms
Elementary Functions
A general exponential function has form y = aebx where a and b are
constants and the base of the exponential has been chosen to be e.
Occasionally we have an exponential function with a different base and
need to change the function into this general form. Here are some
examples.
Part 3, Exponential Functions & Logarithms
Lecture 3.4b, Working With Logarithms, continued
Dr. Ken W. Smith
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Simplification of logarithms
Worked problems on general exponential form.
Write the following functions in the form y = aebx .
1 y = 2x
2 y = 22x−2
3 y = 3x
4 y = 10x+1
Solutions.
1 Since 2 = eln 2 then y = 2x = (eln 2 )x = e(ln 2)x . (Here a = 1 and
b = ln 2.)
22x
1
2 By basic properties of exponents, 22x−2 = 2 = ( 4 )22x . Since
2
2 = eln 2 then 22x = (eln 2 )2x = e(ln 2)(2x) = e(2 ln 2) x . So our answer
is
3
4
1 2(ln 2)x
4e
. (Here a =
1
4
and b = 2 ln 2.)
Since 3 = eln 3 then y = 3x = (eln 3 )x = e(ln 3)x . (Here a = 1 and
b = ln 3.)
By
our basic properties of Elementary
exponents,
10x+1 = 10x · 10. Since2013 15 / 23
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Functions
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Understanding logarithms
Suppose you do not have a calculator. You are asked to compute the
logarithms, base 10, of the first ten positive integers, 1,2,3, . . . , 10. That
is, you are asked to fill out as much of the following table as possible.
log(1)
log(2)
log(3)
log(4)
log(5)
log(6)
log(7)
log(8)
log(9)
log(10)
Here are a series of questions designed to explore what we know about
logarithms.
1 Which of these ten logarithms can you compute immediately, without
any
further
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Understanding logarithms
Understanding logarithms
Solution.
1
2
Suppose I tell you that log(2) = 0.30103. Which of these ten
logarithms can you compute now?
3
Suppose I tell you that log(2) = 0.30103 and log(3) = 0.47712. Now
which of these ten logarithms can you compute?
4
Given the information for log(2) and log(3), fill out as much of the
table as possible.
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Understanding logarithms
Using properties of logs, without any further information, we know
that log(1) = 0 and log(10) = 1. So we can fill in the first and last
line:
log(1) 0
log(2)
log(3)
log(4)
log(5)
log(6)
log(7)
log(8)
log(9)
log(10) 1
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Understanding logarithms
So now we know:
2
log(1)
log(2)
log(3)
log(4)
log(5)
log(6)
log(7)
log(8)
log(9)
log(10)
Using log(2) = .30103, we have
log(4) = log(22 ) = 2 log 2 = 2(0.30103) = 0.60206 and
log(8) = log(23 ) = 3 log 2 = 3(0.30103) = .90309.
We also have
log(5) = log( 10
2 ) = log(10) − log(2) = 1 − .30103 = .69897.
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0
0.30103
0.60206
0.69897
0.90309
1
Elementary Functions
Understanding logarithms
Understanding logarithms
So now we know:
3
log(1)
0
log(2) 0.30103
log(3) 0.47712
log(4) 0.60206
log(5) 0.69897
log(6) 0.77815
log(7)
log(8) 0.90309
log(9) 0.95424
log(10)
1
Using log(2) = .30103 and log(3) = .47712 we have
log(6) = log(3 · 2) = log 2 + log 3 = .30103 + .47712 = .77815.
and log(9) = log(32 ) = 2 log(3) = 2(.47712) = .95424.
(We cannot obtain the log of 7 this way since we cannot write 7 as a
product or quotient of powers of 2 and 3.)
Challenge exercise: with this information on log 2 and log 3 how would you
compute log 7.2?
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Logarithms
In the next presentation we use our log properties to solve a variety of
equations.
(END)
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