2. Common and Natural Logarithms
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Chapter 5
2.
Common and Natural Logarithms
This lesson focuses on the mathematical concepts of exponents, common logarithms, and natural logarithms. It explains how these elements are essential tools for simplifying complex calculations in various fields like engineering, finance, and science. For instance, natural logarithms, which use a special number known as the natural base, are particularly useful in calculating compound interest or understanding exponential growth and decay. Common logarithms, on the other hand, are often employed in tasks like earthquake measurement and audio processing. By understanding these concepts, you can gain a deeper insight into the mathematical principles that govern various phenomena and make problem-solving more efficient.
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Lesson Settings & Tools
| | 14 Theory slides |
| | 14 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Common and Natural Logarithms
Slide of 14
The logarithmic expression log_b a is read
the logarithm of a with base b.Here, the base b is clearly written in the expression. There are two cases in which the base does not need to be written, which will be discussed in this lesson.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Mathematical Operations
The n^(th) root of a real number a expresses another real number that, when multiplied by itself n times, will result in a. Aside from the radical symbol, the notation is made up of the radicand a and the index n.
The resulting number is commonly called a power. In this example, the base 7 is multiplied by itself 4 times. 7^4 = 7 * 7 * 7 * 7_4 However, if the exponent of a number or expression is 1, it does not need to be written. a=a^1 In the case of logarithms, there are two bases that do not need to be written. Take a look at a calculator and try to identify them.
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Discussion
Common Logarithm
A common logarithm is a logarithm of base 10. For example, log_(10) 1000 is called the common logarithm of 1000.
It is equal to 3 because 10^3 is 1000.
Since common logarithms are used so often, the base does not need to be written.
log m ≡ log_(10) m
In the identity above, m is a positive number. Recalling the definition of a logarithm, the common logarithm of m can be defined for positive values of m.
log m=n ⇔ 10^n=m
Common logarithms can be evaluated using a calculator. For example, to evaluate log 34, push LOG, enter 34, and then hit ENTER.
The common logarithm of 34 is about 1.53.
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Evaluating Common Logarithms
Use a calculator to evaluate the common logarithms. Round the answer to two decimal places.
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Example
Earthquakes and Logarithms
A well-known logarithmic scale that is used to express the amount of energy released by an earthquake is the Richter scale.
If two earthquakes have intensity levels I_1 and I_2, and magnitudes on the Richter scale M_1 and M_2, the following formula holds true. log I_1/I_2=M_1-M_2 In 1989, the city of San Francisco was jolted by an earthquake of magnitude 7.1 on the Richter scale. In 1906, nearly 80 years prior, the city had suffered a more powerful earthquake that is estimated to have measured magnitude 7.9 on the same scale. Vincenzo wants to know how many times more intense the 1906 earthquake was than to the 1989 earthquake.
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Hint
Substitute 7.9 and 7.1 for M_1 and M_2, respectively, in the given formula. Then, use the definition of a common logarithm.
Solution
To calculate how many times more intense the 1906 earthquake was compared to the one in 1989, the value of I_1I_2 needs to be found. To do so, 7.9 and 7.1 can be substituted for M_1 and M_2, respectively, in the given formula.
log I_1/I_2=M_1-M_2
SubstituteII
M_1= 7.9, M_2= 7.1
log I_1/I_2= 7.9- 7.1
SubTerm
Subtract term
log I_1/I_2=0.8
This equation can be rewritten by following the definition of a common logarithm. log I_1/I_2=0.8 ⇔ I_1/I_2=10^(0.8) Finally, use a calculator to evaluate 10^(0.8).
The 1906 earthquake was about 6.31 times more intense than the 1989 earthquake.
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Example
Using the Properties of Common Logarithms
After studying the relationship between earthquakes and logarithms, Vincenzo became more interested in this fascinating math topic.
Now he wants to pair the logarithmic expressions that involve common logarithms with their corresponding simplified expression or number. Help him do this!
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Hint
Use the properties of logarithms and the definition of a common logarithm.
Solution
First, simplify the expressions on the left. They can then be paired with their corresponding expressions on the right.
Expression 1
To simplify this expression, the Quotient Property of Logarithms and the definition of a common logarithm will be used.
log 10x - log x
log(m) - log(n)=log(m/n)
log 10x/x
CancelCommonFac
Cancel out common factors
log 10x/x
SimpQuot
Simplify quotient
log 10
log(10) = 1
1
The expression log 10x - log x should be paired with 1.
Expression 2
To simplify this expression, the Power Property of Logarithms, the Product Property of Logarithms, and the definition of a common logarithm will be used.
2log x+log 10x
m*log(a)=log(a^m)
log x^2+log 10x
log(mn)=log(m) + log(n)
log x^2+log 10+log x
log(10) = 1
log x^2+1+log x
CommutativePropAdd
Commutative Property of Addition
1+log x+log x^2
log(m) + log(n)=log(mn)
1+log (x* x^2)
MultBasePow
a*a^m=a^(1+m)
1+log x^3
The expression 2log x+log 10x should be paired with 1+log x^3.
Expression 3
To simplify the third expression, the Product Property of Logarithms, the Quotient Property of Logarithms, a logarithm identity, and the definition of a common logarithm will be used.
log 0.1x
log(mn)=log(m) + log(n)
log 0.1+log x
NumberToFrac
a = 10* a/10
log 10* 0.1/10+log x
Multiply
Multiply
log 1/10+log x
log(m/n)=log(m) - log(n)
log 1-log 10+log x
log(1) = 0
0-log 10+log x
log(10) = 1
0-1+log x
SubTerm
Subtract term
- 1+log x
CommutativePropAdd
Commutative Property of Addition
log x-1
The expression log 0.1x should be paired with log x-1.
Expression 4
To simplify the second to last expression, the Quotient Property of Logarithms, the Power Property of Logarithms, and the definition of a common logarithm will be used.
log 100/x+log x^2
log(m/n)=log(m) - log(n)
log 100- log x+log x^2
WritePow
Write as a power
log 10^2- log x+log x^2
log(a^m)= m*log(a)
2log 10- log x+2log x
log(10) = 1
2(1)- log x+2log x
IdPropMult
Identity Property of Multiplication
2- log x+2log x
AddTerms
Add terms
2+ log x
The expression log 100x+log x^2 should be paired with 2+ log x.
Expression 5
Finally, to simplify the last expression, the Product Property of Logarithms and the Power Property of Logarithms will be used.
log x+log x^2-3log x
log(m) + log(n)=log(mn)
log (x* x^2)-3log x
MultBasePow
a*a^m=a^(1+m)
log x^3-3log x
m*log(a)=log(a^m)
log x^3-log x^3
SubTerm
Subtract term
0
The expression log x+log x^2-3log x should be paired with 0.
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Calculating the Value of an Expression for Bigger Values of n
Consider the following algebraic expression. (1+1/n)^n This expression can be evaluated for different values of n. Pay close attention to the value of the expression as n increases.
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Discussion
The Natural Base
Logarithms and exponents are closely related. This part of the lesson will explore a particular type of exponential expression. Recall the formula for compound interest.
Concept
The Number e
The number e — commonly called the natural base — is an irrational mathematical constant named by the mathematician Leonhard Euler.
e = 2.7182818284...
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Discussion
The Natural Logarithm
A natural logarithm is a logarithm with base e.
Although it is correct to write log_e,
the natural logarithm is more commonly written as ln.
ln m = log_e m
This means that ln m equals the exponent to which e must be raised to equal m.
ln m=n ⇔ m = e^n
Natural logarithms can be evaluated using a calculator. For example, to evaluate ln 21, push LN, input 21, and then hit ENTER.
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Evaluating Natural Logarithms
Use a calculator to evaluate the natural logarithms. Round the answer to two decimal places.
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Example
Using the Properties of Natural Logarithms
Vincenzo already knows how to use the properties of logarithms when dealing with common logarithms. Now he wants to use the properties when natural logarithms are involved.
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Hint
Simplify all the expressions using the properties of logarithms.
Solution
Simplify the expressions so that the expression that is not equivalent with the others can be easily identified.
Expression 1
To simplify this expression, the Product Property of Logarithms, the Quotient Property of Logarithms, and the Power Property of Logarithms will be used.
ln x^2 +ln y/x+ln z
ln(a^b)= b*ln(a)
2ln x +ln y/x+ln z
ln(a/b)=ln(a) - ln(b)
2ln x +ln y-ln x+ln z
SubTerm
Subtract term
ln x +ln y+ln z
ln(a) + ln(b)=ln(ab)
ln xy+ln z
ln(a) + ln(b)=ln(ab)
ln xyz
The expression ln x^2 +ln yx+ln z is equivalent to ln xyz.
Expression 2
To simplify the second expression, the Product Property of Logarithms, the Power Property of Logarithms, and the definition of a natural logarithm will be used.
ln ex+ln y^2z-(ln y+1)
ln(ab)=ln(a) + ln(b)
ln e+ln x+ln y^2+ln z-(ln y+1)
ln(e) = 1
1+ln x+ln y^2+ln z-(ln y+1)
ln(a^b)= b*ln(a)
1+ln x+2ln y+ln z-(ln y+1)
Distr
Distribute - 1
1+ln x+2ln y+ln z-ln y-1
SubTerms
Subtract terms
ln x+ln y+ln z
ln(a) + ln(b)=ln(ab)
ln xy+ln z
ln(a) + ln(b)=ln(ab)
ln xyz
The expression ln ex+ln y^2z-(ln y+1) is also equivalent to ln xyz.
Expression 3
To simplify the third expression, the Quotient Property of Logarithms and the definition of a natural logarithm will be used.
- ln e/xyz+1
ln(a/b)=ln(a) - ln(b)
- (ln e -ln xyz)+1
ln(e) = 1
- (1 -ln xyz)+1
Distr
Distribute - 1
- 1 +ln xyz+1
AddTerms
Add terms
ln xyz
The expression - ln exyz+1 is also equivalent to ln xyz.
Expression 4
To simplify the second to last expression, the Product Property of Logarithms and the Quotient Property of Logarithms will be used.
ln xy+ln z-ln y
ln(a) + ln(b)=ln(ab)
ln xyz-ln y
ln(a) - ln(b)=ln(a/b)
ln xyz/y
CancelCommonFac
Cancel out common factors
ln xyz/y
SimpQuot
Simplify quotient
ln xz
The expression ln xy+ln z-ln y is equivalent to ln xz. Therefore, this expression is not equivalent to the previous expressions.
Expression 5
The last expression will be simplified by using the Power Property of Logarithms and the Product Property of Logarithms.
2ln sqrt(xy) +ln z^2 -ln z
SqrtToPowD
sqrt(a)=a^(12)
2ln (xy)^(12) +ln z^2 -ln z
ln(a^b)= b*ln(a)
2(1/2)ln xy +2ln z -ln z
DenomMultFracToNumber
2 * a/2= a
1ln xy +2ln z -ln z
IdPropMult
Identity Property of Multiplication
ln xy +2ln z -ln z
SubTerm
Subtract term
ln xy +ln z
ln(a) + ln(b)=ln(ab)
ln xyz
The last expression, 2ln sqrt(xy) +ln z^2 -ln z, is also equivalent to ln xyz. Therefore, the expression that is not equivalent to the others is the fourth expression.
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Discussion
Change of Base Formula
Most calculators only calculate common and natural logarithms. These are logarithms with base 10 or e. Luckily, there is a formula that allows any logarithm to be written in terms of common or natural logarithms.
Rule
Change of Base Formula
A logarithm of arbitrary base can be rewritten as the quotient of two logarithms with the same base by using the change of base formula.
log_c a= log_b a/log_b c
This rule is valid for positive values of a,b, and c, where b and c are different than 1.
Proof
Let x=log_c a. Therefore, by the definition of a logarithm, it is known that a=c^x.
x=log_c a ⇔ a=c^x
By the Reflexive Property of Equality, log_b a is equal to itself.
log_b a=log_b a
Substitute
a= c^x
log_b a=log_b c^x
log_b(a^m)= m* log_b(a)
log_b a=xlog_b c
Substitute
x= log_c a
log_b a= log_c alog_b c
DivEqn
.LHS /log_b c.=.RHS /log_b c.
log_b a/log_b c=log_c a
RearrangeEqn
Rearrange equation
log_c a=log_b a/log_b c ✓
Note that this formula is helpful to calculate any logarithm using a calculator, since the new base can be any positive number different than 1. This means that the new base can be 10 or e.
log_c a=log a/log c log_c a=ln a/ln c
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Example
Using the Change of Base Formula
Vincenzo has discovered that he can use the Change of Base Formula to calculate the value of any logarithm.
a Use common logarithms and a calculator to find the value of log_4 23. Round the answer to two decimal places.
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b Use natural logarithms and a calculator to find the value of log_7 56. Round the answer to two decimal places.
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Hint
a By the Change of Base Formula, log_4 23= log 23log 4.
b By the Change of Base Formula, log_7 56= ln 56ln 7.
Solution
a Recall the Change of Base Formula.
log_b a=log_c a/log_c b Here, a, b, and c are positive numbers, with b and c different than 1. While the base c can be any number, in this case it might be most helpful if it was 10. Then the logarithm of 23 with base 4 could be expressed in terms of common logarithms. log_4 23=log 23/log 4 The value of the numeric expression on the right-hand side of the above equation can be found by using a calculator.
File:A2 U5 C2 Example4 2.svg
Therefore, the value of log_4 23 rounded to the nearest hundredth is 2.26.
b Once again, recall the Change of Base Formula, this time with natural logarithms.
ln_b a=log_c a/log_c b Here, a, b, and c are positive numbers, with b and c different than 1. For this exercise, let c be e. Then the logarithm of 56 with base 7 can be expressed in terms of natural logarithms. log_7 56=ln 56/ln 7 The value of the numeric expression on the right-hand side of the above equation can be found by using a calculator.
File:A2 U5 C2 Example4 3.svg
Therefore, the value of log_7 56 rounded to the nearest hundredth is 2.07.
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Closure
Leonhard Euler and His Contributions
In this lesson, the Swiss mathematician Leonhard Euler was mentioned. Euler was also a physicist, astronomer, geographer, logician, and engineer. During his life, Euler came up with principles that set the foundations for most of the mathematics used nowadays. He was a revolutionary thinker in the fields of geometry, calculus, trigonometry, differential equations, and number theory.
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Common and Natural Logarithms
Exercise 2.1
Bizarrap is a young Argentine record producer and DJ who specializes in genres like trap, EDM, Latin trap, and rap.
Suppose that a sound has an intensity of I watts per square meter. Then, the loudness L(I) of this sound in decibels is given by the following formula. L(I)=10log (10^(12)I) In his recording studio, Bizarrap turns up the volume on his latest track so that the intensity of the sound doubles. How many decibels does the loudness increase by? Round the answer to the nearest integer.
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If I is the original intensity and Bizarrap turns up the volume so that the intensity is doubled, the new intensity is 2I. Therefore, the loudness of the sound is now L(2I). L( 2I)=10log (10^(12)( 2I)) We want to calculate how many decibels the loudness increases by. This is the difference between the loudness of the sound with an intensity of 2I and the loudness of the same sound with an intensity of I. L(2I)- L(I) = 10log (10^(12)(2I))- 10log (10^(12)I) Let's calculate this difference by applying properties of logarithms. First, note that both arguments consist of a product. Therefore, we can use the Product Property of Logarithms to expand them.
Next, we can use the Distributive Property to distribute 10 and - 10.
Now we can simplify the expression.
Note that we have a subtraction of two logarithms with the same base. This means that we can use the Quotient Property of Logarithms to further simplify the argument.
The loudness increases by 10log 2 decibels. Since we have a common logarithm, we can calculate the numerical value of this expression by using a calculator.
Therefore, rounded to the nearest integer, the loudness increases by 3 decibels.
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Solution
The loudness L(I) in decibels of a sound with an intensity of I watts per square meter is given by the following formula. L(I)=10log (10^(12)I) The record producer Bizarrap is recording a session with the Argentine artist Nicki Nicole.
The intensity of the sound of this session is 10 times greater than the intensity I of Bizarrap's previous session. How many decibels did the loudness increase by?
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We know that I is the intensity of the previous session and that the intensity of the session with Nicki Nicole is 10 times greater. Therefore, the intensity of this session is 10I. This means that its loudness is L(10I). L( 10I)=10log (10^(12)( 10I)) We want to calculate how many decibels the loudness increased by. This is the difference between the loudness of the session with Nicki Nicole, with an intensity of 10I, and the loudness of the previous session, which had an intensity of I. L(10I)- L(I) = 10log (10^(12)(10I))- 10log (10^(12)I) Let's calculate this difference by applying properties of logarithms. First, since we have a product in each argument, we can apply the Product Property of Logarithms to expand the expression.
Now let's use the Distributive Property to distribute 10 and - 10 and simplify.
Finally, since we have a subtraction of two logarithms with the same base, we can use the Quotient Property of Logarithms to further simplify the expression.
Recall that a common logarithm is a logarithm with base 10. This means that log 10=1.
Therefore, the loudness increased by 10 decibels.
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Solution
The loudness L(I) in decibels of a sound with an intensity of I watts per square meter is given by the following formula. L(I)=10log (10^(12)I) Certain types of whale can make a sound with an intensity that is one million times greater than the intensity of the loudest sound a human being can make.
External credits: @brgfx
Calculate the difference in the decibel levels of the sounds made by this type of whale and a human being.
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Let I be the intensity of the loudest sound a human can make. Since the whale can produce sound with an intensity that is 1 000 000 times greater than I, the intensity of the whale's sound is 1 000 000I. Therefore, the loudness of this sound is L(1 000 000I). L( 1 000 000I)=10log (10^(12)( 1 000 000I)) We want to calculate the difference in the decibel levels of the sounds made by this certain type of whale and a human. This means that we need to find L(1 000 000I)-L(I). L(1 000 000I)- L(I) = 10log (10^(12)(1 000 000I))- 10log (10^(12)I) Let's calculate this difference by applying properties of logarithms. Notice that we have a product in both arguments, so we can start by using the Product Property of Logarithms to expand the expressions.
Let's simplify the expression by distributing 10 and - 10.
Note that now we have a subtraction of two logarithms with the same base. This means that we can use the Quotient Property of Logarithms to further simplify the resulting single logarithm.
To calculate this logarithmic expression, we will rewrite 1 000 000 as 10^6 and apply the Power Property of Logarithms. Then, we will use the fact that log 10=1. Let's do it!
We found that the difference of the sounds made by this type of whale and a human being is 60 decibels.
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Under certain conditions, the wind speed w knots at an altitude of h meters above a plane surface can be modeled by the following function. w(h)=2ln 100h If the altitude is doubled, by what amount does the wind speed increase? Round the answer to three decimal places.
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If the altitude h is doubled, the new altitude is 2h. This means that the wind speed that corresponds to this altitude is w(2h). w( 2h)=2ln (100( 2h)) To find the increase in the wind speed, we need to find the difference of w(2h) and w(h). w(2h)- w(h) = 2ln (100(2h))- 2ln 100h Let's use the properties of logarithms to calculate this difference. First, since we have a product in both arguments, we can use the Product Property of Logarithms to expand the expressions.
Let's now simplify the expression by distributing 2 and - 2.
Finally, because we have the difference of two logarithms with the same base, we can use the Quotient Property of Logarithms to further simplify the expression.
We ended with an expression that contains a natural logarithm. Let's calculate its value by using a calculator.
Therefore, if the altitude is doubled, the wind speed increases by about 1.386 knots.
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Solution
An erg is a unit of energy that is equal to 10^(- 7) joules. The amount of energy E ergs that an earthquake releases can be related to its magnitude M on the Richter scale by the following equation. log E=11.8+1.5M Use this equation to calculate the amount of energy released in ergs by the earthquake that hit Chile in 1960, which measured 8.5 on the Richter scale.
Write the answer in scientific notation rounded to three significant figures.
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To calculate the amount of energy released by the earthquake that occurred in Chile in 1960, let's substitute M=8.5 into the given equation.
Next, to find the value of E, we will use the definition of a logarithm. Recall that a common logarithm is a logarithm with base 10. Definition:& log_b a= c ⇔ b^c= a Equation:& log E= 24.55 ⇔ 10^(24.55)= E Therefore, E=10^(24.55) ergs. Let's write this value in scientific notation and round to three significant figures.
The amount of energy released by the earthquake is about 3.55* 10^(24) ergs.
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Common and Natural Logarithms
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