Mathematics

Natural Logarithm

The natural logarithm, denoted as ln, is a mathematical function that represents the logarithm to the base of the mathematical constant e, approximately equal to 2.718. It is commonly used in various mathematical and scientific applications, particularly in calculus, to solve exponential growth and decay problems, and to simplify complex mathematical expressions involving exponential functions.

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  • Book cover image for: Intermediate Algebra
    eBook - PDF

    Intermediate Algebra

    A Guided Approach

    • Rosemary Karr, Marilyn Massey, R. Gustafson, , Rosemary Karr, Marilyn Massey, R. Gustafson(Authors)
    • 2014(Publication Date)
    We have seen that the logarithm of a number is an exponent. For Natural Logarithms, ln x is the exponent to which e is raised to obtain x . In equation form, we write e ln x 5 x To find the base-e logarithms of numbers, we can use a calculator. 1 Section 2 3 Evaluate a Natural Logarithm. Solve a logarithmic equation with a calculator. Graph a Natural Logarithmic function. Solve an application involving a Natural Logarithm. 1 Natural Logarithm 4 Evaluate each expression. 1. log 4 16 2. log 2 1 8 3. log 5 5 4. log 7 1 9.4 Vocabulary Getting Ready Objectives Natural Logarithms John Napier 1550–1617 Napier is famous for his work with Natural Logarithms. In fact, Natural Logarithms are often called Napierian logarithms . He also invented a device, called Napier’s rods , that did multiplica-tions mechanically. His device was a forerunner of modern-day computers. Unless otherwise noted, all content on this page is © Cengage Learning. Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 9.4 Natural Logarithms 651 Use a calculator to find each value. a. ln 17.32 b. ln 1 log 0.05 2 a. We can enter these numbers and press these keys: Scientific Calculator Graphing Calculator 17.32 LN LN 17.32 ) ENTER Either way, the result is 2.851861903. b. We can enter these numbers and press these keys: Scientific Calculator Graphing Calculator 0.05 LOG LN LN ( LOG 0.05 ) ) ENTER Either way, we obtain an error, because log 0.05 is a negative number. Because the domain of ln x is 1 0, ` 2 , we cannot take the logarithm of a negative number.
  • Book cover image for: Explorations in College Algebra
    • Linda Almgren Kime, Judith Clark, Beverly K. Michael(Authors)
    • 2018(Publication Date)
    • Wiley
      (Publisher)
    To solve this equation, we need Natural Logarithms that use e as a base. The Natural Logarithm The common logarithm uses 10 as a base. The Natural Logarithm uses e as a base and is written ln x rather than log e x. Scientific calculators have a key that computes ln x. The Natural Logarithm The logarithm base e of x is the exponent of e needed to produce x. Logarithms base e are called Natural Logarithms and are written as ln x. ln x = c means that e c = x (x > 0) The properties for Natural Logarithms (base e) are similar to the properties for common loga- rithms (base 10). Like the common logarithm, ln A is not defined when A ≤ 0. 6.2 Using Natural Logarithms to Solve Exponential Equations Base e 345 Assume A and B are positive real numbers and p is any real number. Rules of Common Logarithms 1. log(A ⋅ B) = log A + log B 3. log A p = p log A 2. log(A/B) = log A − log B 4. log 1 = 0 (since 10 0 = 1) Rules of Natural Logarithms 1. ln(A ⋅ B) = ln A + ln B 3. ln A p = p ln A 2. ln(A/B) = ln A − ln B 4. ln 1 = 0 (since e 0 = 1) We can use the rules of Natural Logarithms to manipulate expressions involving natural logs. EXAMPLE 1 Expanding Expressions with Natural Logs Expand, using the laws of logarithms, the expression: x x ln 3 2 + − . Solution Rewrite using exponents x x x x ln 3 2 ln 3 2 1/2 + − = + −       Rule 3 of ln x x 1 2 ln 3 2 = + −       Rule 2 of ln x x 1 2 [ln( 3) ln( 2)] = + − − EXAMPLE 2 Contracting Expressions with Natural Logs Contract, expressing the answer as a single logarithm: x x ln( 1 ) ln( 1) 1 3 1 3 − + + .
  • Book cover image for: Functions Modeling Change
    eBook - PDF

    Functions Modeling Change

    A Preparation for Calculus

    • Eric Connally, Deborah Hughes-Hallett, Andrew M. Gleason(Authors)
    • 2019(Publication Date)
    • Wiley
      (Publisher)
    The log base  is used so frequently that it has its own notation: ln , read as the natural log of . We make the following definition: For  > 0, ln  is the power of  that gives  or, in symbols,  = ln  means   = , and  is called the Natural Logarithm of . 5.1 LOGARITHMS AND THEIR PROPERTIES 167 Just as the functions 10  and log  are inverses, so are   and ln . The function ln  has similar properties to the common log function: Properties of the Natural Logarithm • By definition,  = ln  means  =   . • In particular, ln 1 = 0 and ln  = 1. • The functions   and ln  are inverses, so they “undo” each other: ln(  ) =  for all   ln  =  for  > 0. • For  and  both positive and any value of , ln() = ln  + ln  ln (   ) = ln  − ln  ln(  ) =  ⋅ ln . Example 6 Solve for : (a) 5 2 = 50 (b) 3  = 100 (c) ln( + 2) = 3 Solution (a) We first divide both sides by 5 to obtain  2 = 10. Taking the natural log of both sides, we have ln( 2 ) = ln 10 2 = ln 10  = ln 10 2 = 1.151. (b) Taking natural logs of both sides yields ln(3  ) = ln 100  ln 3 = ln 100  = ln 100 ln 3 = 4.192. (c) If  = ln  then   = , so since ln( + 2) = 3, we know  + 2 =  3  =  3 − 2. In part (a), using the natural log, rather than log base 10, simplified the calculation because the base was . In part (b), we used natural logs, but we could equally well have used logs to base 10. Misconceptions and Calculator Errors Involving Logs It is important to know how to use the properties of logarithms. It is equally important to recognize statements that are not true. Beware of the following: 168 Chapter 5 LOGARITHMIC FUNCTIONS • log( + ) is not the same as log  + log  • log( − ) is not the same as log  − log  • log() is not the same as (log )(log ) • log (   ) is not the same as log  log  • log ( 1  ) is not the same as 1 log  . There are no formulas to simplify either log( + ) or log( − ).
  • Book cover image for: Calculus, Volume 1
    • Tom M. Apostol(Author)
    • 2019(Publication Date)
    • Wiley
      (Publisher)
    6.2 Motivation for the definition of the Natural Logarithm as an integral The logarithm is an example of a mathematical concept that can be defined in many different ways. When a mathematician tries to formulate a definition of a concept, such as the logarithm, he usually has in mind a number of properties he wants this concept to have. By examining these properties, he is often led to a simple formula or process that might serve as a definition from which all the desired properties spring forth as logical deductions. We shall illustrate how this procedure may be used to arrive at the definition of the logarithm which is given in the next section. One of the properties we want logarithms to have is that the logarithm of a product should be the sum of the logarithms of the individual factors. Let us consider this property by itself and see where it leads us. If we think of the logarithm as a function f , then we want this function to have the property expressed by the formula f (xy) = f (x) + f (y) (6.4) whenever x, y, and xy are in the domain of f . An equation like (6.4), which expresses a relationship between the values of a function at two or more points, is called a functional equation. Many mathematical problems can be reduced to solving a functional equation, a solution being any function which satisfies the equation. Ordinarily an equation of this sort has many different solutions, and it is usually very difficult to find them all. It is easier to seek only those solutions which have some additional property such as continuity or differentiability. For the most part, these are the only solutions we are interested in anyway. We shall adopt this point of view and determine all differentiable solutions of (6.4). But first let us try to deduce what information we can from (6.4) alone, without any further restrictions on f . One solution of (6.4) is the function that is zero everywhere on the real axis.
  • Book cover image for: College Algebra
    eBook - PDF
    Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. Chapter 5 Exponential and Logarithmic Functions 528 Solve: log x 5 1.87737. Give the result to four decimal places. Now Try Exercise 81. 3. Evaluate Natural Logarithms We have seen the importance of the number e in mathematical models of events in nature. Base-e logarithms are just as important. They are called Natural Logarithms or Napierian logarithms after John Napier (1550–1617). They are usually written as ln x, rather than log e x: ln x means log e x Like all logarithmic functions, the domain of f sxd 5 ln x is the interval s0, `d, and the range is the interval s2`, `d. To estimate the base-e logarithms of numbers, we can use a calculator. Scien- tific and graphing calculators have a Natural Logarithm key LN LN , which is used like the LOG LOG key. Self Check 5 Accent on Technology Approximating Natural Logarithms We can use the LN LN key on a graphing calculator to evaluate base-e logarithms. Figure 5-18 shows the evaluation of ln 2.34 and demonstrates that a logarithm is an exponent. Notice that the logarithm is evaluated, and then the answer is used as the exponent of e to get 2.34. Remember: ln 2.34 is the exponent of e that will yield 2.34. FIGURE 5-18 Using a Calculator to Approximate Natural Logarithms Use a calculator to find a. ln 17.32 b. lnslog 0.05d a. ln 17.32 < 2.851861903, which means e 2.851861903 < 17.32 b. lnslog 0.05d is undefined because logs0.05d , 0. Our calculator gives an error message. SOLUTION EXAMPLE 6 Find each value to four decimal places: a. ln H9266 b. ln 1 log 1 3 2 Now Try Exercise 75. Self Check 6 Using a Calculator to Solve Natural Logarithmic Equations Solve each equation and give the result to four decimal places: a. ln x 5 1.335 b. ln x 5 log 5.5 EXAMPLE 7 Copyright 2017 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
  • Book cover image for: Algebra
    eBook - PDF

    Algebra

    Form and Function

    • William G. McCallum, Eric Connally, Deborah Hughes-Hallett(Authors)
    • 2015(Publication Date)
    • Wiley
      (Publisher)
    In other words, if  = ln  then   =  and if   =  then  = ln . Example 2 Solve 12 −0.1 = 3 using the Natural Logarithm function. Solution Since 12 −0.1 = 3, we have  −0.1 = 0.25. So we need to find the power of  that is equal to 0.25. By definition of the Natural Logarithm function, the exponent we need is ln 0.25, so −0.1 = ln 0.25 by definition of ln function 276 Chapter 7 LOGARITHMS  = −10 ln 0.25 multiply by −10 = 13.863. using a calculator Nearly all scientific and graphing calculators have an ln button to find the natural log. The answer agrees with our solution in Example 1, where we solved the same equation to obtain  = −10 ⋅ log 0.25 log  = 13.863. Properties of Natural Logarithms The Natural Logarithm function ln  has similar properties to the common log function log : Properties of the Natural Logarithm • The operation of raising  to a power and the ln operation undo each other: ln(  ) =  for all   ln  =  for  > 0 • For  and  both positive and any value of , ln() = ln  + ln  ln (   ) = ln  − ln  ln(  ) =  ⋅ ln  • Since  0 = 1, ln 1 = 0. Using Natural Logarithms to Find Continuous Growth Rates In Example 1 on page 244, we saw how to determine the growth factor  and the percent growth rate  of an exponential function like  = 5000 0.147 . Using the Natural Logarithm, we can determine the continuous growth rate if we know the growth factor. Example 3 The value of an investment in year  is given by  = 1000(1.082)  . Give the initial value, the annual growth factor, the percent growth rate, and the continuous growth rate. Solution The initial value is  = $1000, the growth factor is  = 1.082, and the percent growth rate is  =  − 1 = 0.082 = 8.2%.
  • Book cover image for: Precalculus
    eBook - PDF

    Precalculus

    A Prelude to Calculus

    • Sheldon Axler(Author)
    • 2016(Publication Date)
    • Wiley
      (Publisher)
    Properties of powers of e e 0 = 1 e 1 = e e x e y = e x+y e -x = 1 e x e x e y = e x-y (e x ) y = e xy The Natural Logarithm of a positive number x, denoted ln x, equals log e x. Thus Because the function ln x is the inverse of the function e x , the graph of ln x could be obtained by flipping the graph above across the line y = x. the graph of the Natural Logarithm looks similar to the graphs of the functions log 2 x or log x or log b x for any number b > 1. Specifically, ln x grows slowly as x gets large. Furthermore, if x is a small positive number, then ln x is a negative number with large absolute value, as shown in the following figure. x y The graph of ln x on the interval [e -2 , e 2 ]. The same scale is used on both axes to show the slow growth of ln x and the rapid descent near 0 toward negative numbers with large absolute value. 2 8 Chapter 3 Exponential Functions, Logarithms, and e The domain of ln x is the set of positive numbers. The range of ln x is the set of real numbers. Furthermore, ln x is an increasing function because it is the inverse of the increasing function e x . Because the Natural Logarithm is the logarithm with base e, it has all the properties In this book, as in most precalculus books, log x means log 10 x. However, the Natural Logarithm is so important that many mathematicians (and WolframAlpha) use log x to denote the Natural Logarithm rather than the logarithm with base 10. we saw earlier for logarithms with any base. For review, we summarize the key properties here. In the box below, we assume x and y are positive numbers. Properties of the Natural Logarithm ln 1 = 0 ln e = 1 ln( xy) = ln x + ln y ln 1 x = - ln x ln x y = ln x - ln y ln( x t ) = t ln x The exponential function e x and the Natural Logarithm ln x (which equals log e x) are the inverse functions for each other, just as the functions 2 x and log 2 x are the inverse functions for each other.
  • Book cover image for: Algebra and Trigonometry
    • Sheldon Axler(Author)
    • 2011(Publication Date)
    • Wiley
      (Publisher)
    the name natural. No base for logarithms other than e produces such a nice approximation formula. Approximation of the Natural Logarithm If |t | is small, then ln(1 + t) ≈ t . Consider now the figure below, where we assume that t is positive but not necessarily small. In this figure, ln(1 + t) equals the area of the yellow region under the curve. y  1x 1 1t x 1 y The area of the yellow region under the curve is greater than the area of the lower rectangle and is less than the area of the large rectangle. The yellow region above contains the lower rectangle; thus the lower rectangle has a smaller area. The lower rectangle has base t and height 1 1+t and hence has area t 1+t . Thus t 1 + t < ln(1 + t). The large rectangle in the figure above has base t and height 1 and thus has area t . The yellow region above is contained in the large rectangle; thus the large rectangle has a bigger area. In other words, ln(1 + t) < t. Putting together the inequalities from the previous two paragraphs, we have the result below. This result is valid for all positive numbers t, regardless of whether t is small or large. Inequalities with the Natural Logarithm If t > 0, then t 1 + t < ln(1 + t) < t . If t is small, then t 1+t and t are close to each other, showing that either one is a good estimate for ln(1 + t). For small t , the estimate ln(1 + t) ≈ t is usually easier to use than the estimate ln(1 + t) ≈ t 1+t . However, if we need an estimate that is either slightly too large or slightly too small, then the result above shows which one to use. 368 chapter 6 e and the Natural Logarithm Approximations with the Exponential Function Now we turn to approximations of e x . example 2 Discuss the behavior of e x for |x| a small number. 0. 05 0. 1 0. 05 0. 1 x 0. 9 1 1. 1 y The graphs of y = e x (blue) and y = 1 + x (red) on [-0.1, 0.1]. To save space, the horizontal axis has been drawn at y = 0.85 instead of at the usual location y = 0.
  • Book cover image for: College Algebra
    eBook - PDF
    • Sheldon Axler(Author)
    • 2011(Publication Date)
    • Wiley
      (Publisher)
    the name natural. No base for logarithms other than e produces such a nice approximation formula. Approximation of the Natural Logarithm If |t | is small, then ln(1 + t) ≈ t . Consider now the figure below, where we assume that t is positive but not necessarily small. In this figure, ln(1 + t) equals the area of the yellow region under the curve. y  1x 1 1t x 1 y The area of the yellow region under the curve is greater than the area of the lower rectangle and is less than the area of the large rectangle. The yellow region above contains the lower rectangle; thus the lower rectangle has a smaller area. The lower rectangle has base t and height 1 1+t and hence has area t 1+t . Thus t 1 + t < ln(1 + t). The large rectangle in the figure above has base t and height 1 and thus has area t . The yellow region above is contained in the large rectangle; thus the large rectangle has a bigger area. In other words, ln(1 + t) < t. Putting together the inequalities from the previous two paragraphs, we have the result below. This result is valid for all positive numbers t, regardless of whether t is small or large. Inequalities with the Natural Logarithm If t > 0, then t 1 + t < ln(1 + t) < t . If t is small, then t 1+t and t are close to each other, showing that either one is a good estimate for ln(1 + t). For small t , the estimate ln(1 + t) ≈ t is usually easier to use than the estimate ln(1 + t) ≈ t 1+t . However, if we need an estimate that is either slightly too large or slightly too small, then the result above shows which one to use. 368 chapter 6 e and the Natural Logarithm Approximations with the Exponential Function Now we turn to approximations of e x . example 2 Discuss the behavior of e x for |x| a small number. 0.05 0.1 0.05 0.1 x 0.9 1 1.1 y The graphs of y = e x (blue) and y = 1 + x (red) on [-0.1, 0.1]. To save space, the horizontal axis has been drawn at y = 0.85 instead of at the usual location y = 0.
  • Book cover image for: Precalculus: Mathematics for Calculus, International Metric Edition
    • James Stewart, Lothar Redlin, Saleem Watson(Authors)
    • 2016(Publication Date)
    Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 358 CHAPTER 4 ■ Exponential and Logarithmic Functions EXAMPLE 5 ■ Evaluating Logarithms with the Change of Base Formula Use the Change of Base Formula and common or Natural Logarithms to evaluate each logarithm, rounded to five decimal places. (a) log 8 5 (b) log 9 20 SOLUTION (a) We use the Change of Base Formula with b  8 and a  10: log 8 5  log 10 5 log 10 8 < 0.77398 (b) We use the Change of Base Formula with b  9 and a  e: log 9 20  ln 20 ln 9 < 1.36342 Now Try Exercises 59 and 61 ■ EXAMPLE 6 ■ Using the Change of Base Formula to Graph a Logarithmic Function Use a graphing calculator to graph f 1 x 2  log 6 x . SOLUTION Calculators don’t have a key for log 6 , so we use the Change of Base For- mula to write f 1 x 2  log 6 x  ln x ln 6 Since calculators do have an LN key, we can enter this new form of the function and graph it. The graph is shown in Figure 1. Now Try Exercise 67 ■ CONCEPTS 1. The logarithm of a product of two numbers is the same as the of the logarithms of these numbers. So log 5 1 25 # 125 2   . 2. The logarithm of a quotient of two numbers is the same as the of the logarithms of these numbers. So log 5 A 25 125 B   . 3. The logarithm of a number raised to a power is the same as the times the logarithm of the number. So log 5 1 25 10 2  # . 4. We can expand log a x 2 y z b to get . 5. We can combine 2 log x  log y  log z to get . 6. (a) Most calculators can find logarithms with base and base . To find logarithms with different bases, we use the Formula. To find log 7 12, we write log 7 12  log log < (b) Do we get the same answer if we perform the calculation in part (a) using ln in place of log? 7–8 ■ True or False? 7. (a) log1 A  B 2 is the same as log A  log B. (b) log AB is the same as log A  log B. 8. (a) log A B is the same as log A  log B. (b) log A log B is the same as log A  log B.
  • Book cover image for: College Algebra
    eBook - PDF

    College Algebra

    Building Skills and Modeling Situations

    • Charles P. McKeague, Katherine Yoshiwara, Denny Burzynski(Authors)
    • 2013(Publication Date)
    • XYZ Textbooks
      (Publisher)
    e = lim x → ∞  1 + 1 _ x  x This notation indicates that e is the number that is approached by the expression  1 + 1 _ x  x as x becomes larger and larger. To visualize this process, we can build a table. 0 10 500 1,200 FIGURE 9 422 Chapter 6 Exponential and Logarithmic Functions By making x large enough, we can approximate e to as many decimal places as we like. The number e to 9 decimal places is 2.718281828. For the work we are going to do with the number e, we only need to know that it is an irrational number that is approximately 2.7183. Here are a table and graph (Figure 10) for the natural exponential function y = f (x) = e x One common application of natural exponential functions is with interest-bearing accounts. In Example 5, we worked with the formula A(t) = P  1 + r _ n  nt x  1 + 1 __ x  x 1  1 + 1 _ 1  1 = 2 10  1 + 1 __ 10  10 = (1.1) 10 = 2.6 10 2  1 + 1 _ 100  10 2 = (1.01) 10 2 = 2.70 10 3  1 + 1 _ 1,000  10 3 = (1.001) 10 3 = 2.717 10 4  1 + 1 _ 10,000  10 4 = (1.0001) 10 4 = 2.7181 10 5  1 + 1 _ 100,000  10 5 = (1.00001) 10 5 = 2.71827 10 6  1 + 1 _ 1,000,000  10 6 = (1.000001) 10 6 = 2.718280 10 7  1 + 1 _ 10,000,000  10 7 = (1.0000001) 10 7 = 2.7182817 ↓ ∞ ↓ e FIGURE 10 –3 –2 –1 1 2 3 –1 1 2 3 4 5 y = e x -3 -2 -1 1 2 3 -1 1 2 3 4 5 x y y = e x x f (x) = e x −2 f (−2) = e −2 = 1 __ e 2 ≈ 0.135 −1 f (−1) = e −1 = 1 _ e ≈ 0.368 0 f (0) = e 0 = 1 1 f (1) = e 1 = e ≈ 2.72 2 f (2) = e 2 ≈ 7.39 3 f (3) = e 3 ≈ 20.09 6.1 Exponential Functions 423 that gives the amount of money in an account if P dollars are deposited for t years at annual interest rate r, compounded n times per year. In Example 5, the number of compounding periods was four. What would happen if we let the number of com- pounding periods become larger and larger, so that we compounded the interest every day, then every hour, then every second, and so on? If we take this as far as it can go, we end up compounding the interest every moment.
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