1.5: Exponential and Logarithmic Functions (2024)

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    • 1.5: Exponential and Logarithmic Functions (1)
    • Gilbert Strang & Edwin “Jed” Herman
    • OpenStax

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    Learning Objectives
    • Identify the form of an exponential function.
    • Explain the difference between the graphs of \(x^{b}\) and \(b^{x}\).
    • Recognize the significance of the number \(e\).
    • Identify the form of a logarithmic function.
    • Explain the relationship between exponential and logarithmic functions.
    • Describe how to calculate a logarithm to a different base.
    • Identify the hyperbolic functions, their graphs, and basic identities.

    In this section we examine exponential and logarithmic functions. We use the properties of these functions to solve equations involving exponential or logarithmic terms, and we study the meaning and importance of the number \(e\). We also define hyperbolic and inverse hyperbolic functions, which involve combinations of exponential and logarithmic functions. (Note that we present alternative definitions of exponential and logarithmic functions in the chapter Applications of Integrations, and prove that the functions have the same properties with either definition.)

    Exponential Functions

    Exponential functions arise in many applications. One common example is population growth. For example, if a population starts with \(P_0\) individuals and then grows at an annual rate of \(2\%\), its population after 1 year is

    \[P(1)=P_0+0.02P_0=P_0(1+0.02)=P_0(1.02).\nonumber \]

    Its population after 2 years is

    \[P(2)=P(1)+0.02P(1)=P(1)(1.02)=P_0(1.02)^2.\nonumber \]

    In general, its population after \(t\) years is

    \[P(t)=P_0(1.02)^t,\nonumber \]

    which is an exponential function. More generally, any function of the form \(f(x)=b^x\), where \(b>0\), \(b≠1\), is an exponential function with base \(b\) and exponent \(x.\) Exponential functions have constant bases and variable exponents. Note that a function of the form \(f(x)=x^b\) for some constant \(b\) is not an exponential function but a power function.

    To see the difference between an exponential function and a power function, we compare the functions \(y=x^2\) and \(y=2^x\). In Table \(\PageIndex{1}\), we see that both \(2^x\) and \(x^2\) approach infinity as \(x→∞\). Eventually, however, \(2^x\) becomes larger than \(x^2\) and grows more rapidly as \(x→∞\). In the opposite direction, as \(x→−∞\), \(x^2→∞\), whereas \(2^x→0\). The line \(y=0\) is a horizontal asymptote for \(y=2^x\).

    Table \(\PageIndex{1}\)
    \(x\) -3 -2 -1 0 1 2 3 4 5 6
    \(x^2\) 9 4 1 0 1 4 9 16 25 36
    \(2^x\) 1/8 1/4 1/2 1 2 4 8 16 32 64

    In Figure \(\PageIndex{1}\), we graph both \(y=x^2\) and \(y=2^x\) to show how the graphs differ.

    1.5: Exponential and Logarithmic Functions (2)

    Evaluating Exponential Functions

    Recall the properties of exponents: If \(x\) is a positive integer, then we define \(b^x=b⋅b⋯b\) (with \(x\) factors of \(b\)). If \(x\) is a negative integer, then \(x=−y\) for some positive integer \(y\), and we define \(b^x=b^{−y}=1/b^y\). Also, \(b^0\) is defined to be \(1\). If \(x\) is a rational number, then \(x=p/q\), where \(p\) and \(q\) are integers and \(b^x=b^{p/q}=\sqrt[q]{b^p}\). For example, \(9^{3/2}=\sqrt{9^3}=\left(\sqrt{9}\right)^3=27\). However, how is \(b^x\) defined if \(x\) is an irrational number? For example, what do we mean by \(2^{\sqrt{2}}\)? This is too complex a question for us to answer fully right now; however, we can make an approximation.

    Table \(\PageIndex{2}\): Values of \(2^x\) for a List of Rational Numbers Approximating \(\sqrt{2}\)
    \(x\) 1.4 1.41 1.414 1.4142 1.41421 1.414213
    \(2^x\) 2.639 2.65737 2.66475 2.665119 2.665138 2.665143

    In Table \(\PageIndex{2}\), we list some rational numbers approaching \(\sqrt{2}\), and the values of \(2^x\) for each rational number \(x\) are presented as well. We claim that if we choose rational numbers \(x\) getting closer and closer to \(\sqrt{2}\), the values of \(2^x\) get closer and closer to some number \(L\). We define that number \(L\) to be \(2^{\sqrt{2}}\).

    Example \(\PageIndex{1}\): Bacterial Growth

    Suppose a particular population of bacteria is known to double in size every \(4\) hours. If a culture starts with \(1000\) bacteria, the number of bacteria after \(4\) hours is \(n(4)=1000⋅2\). The number of bacteria after \(8\) hours is \(n(8)=n(4)⋅2=1000⋅2^2\). In general, the number of bacteria after \(4m\) hours is \(n(4m)=1000⋅2^m\). Letting \(t=4m\), we see that the number of bacteria after t hours is \(n(t)=1000⋅2^{t/4}\). Find the number of bacteria after \(6\) hours, \(10\) hours, and \(24\) hours.

    Solution

    The number of bacteria after 6 hours is given by

    \[n(6)=1000⋅2^{6/4}≈2828\, \text{bacteria}. \nonumber \]

    The number of bacteria after \(10\) hours is given by

    \[n(10)=1000⋅2^{10/4}≈5657\, \text{bacteria}. \nonumber \]

    The number of bacteria after \(24\) hours is given by \(n(24)=1000⋅2^6=64,000\) bacteria.

    Exercise \(\PageIndex{1}\)

    Given the exponential function \(f(x)=100⋅3^{x/2}\), evaluate \(f(4)\) and \(f(10)\).

    Answer

    \(f(4)=900\)

    \(f(10)=24,300\).

    Graphing Exponential Functions

    For any base \(b>0\), \(b≠1\), the exponential function \(f(x)=b^x\) is defined for all real numbers \(x\) and \(b^x>0\). Therefore, the domain of \(f(x)=b^x\) is \((−∞,∞)\) and the range is \((0,∞)\). To graph \(b^x\), we note that for \(b>1\), \(b^x\) is increasing on \((−∞,∞)\) and \(b^x→∞\) as \(x→∞\), whereas \(b^x→0\) as \(x→−∞\). On the other hand, if \(0<b<1\), \(f(x)=b^x\) is decreasing on \((−∞,∞)\) and \(b^x→0\) as \(x→∞\) whereas \(b^x→∞\) as \(x→−∞\) (Figure \(\PageIndex{2}\)).

    1.5: Exponential and Logarithmic Functions (3)

    Note that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.

    Laws of Exponents

    For any constants \(a>0\), \(b>0\), and for all \(x\) and \(y,\)

    1. \[b^x⋅b^y=b^{x+y} \nonumber \]
    2. \[\dfrac{b^x}{b^y}=b^{x−y} \nonumber \]
    3. \[(b^x)^y=b^{xy} \nonumber \]
    4. \[(ab)^x=a^xb^x \nonumber \]
    5. \[\dfrac{a^x}{b^x}=\left(\dfrac{a}{b}\right)^x \nonumber \]
      Example \(\PageIndex{2}\): Using the Laws of Exponents

      Use the laws of exponents to simplify each of the following expressions.

      1. \(\dfrac{(2x^{2/3})^3}{(4x^{−1/3})^2}\)
      2. \(\dfrac{(x^3y^{−1})^2}{(xy^2)^{−2}}\)

      Soution

      a. We can simplify as follows:

      \[\dfrac{(2x^{2/3})^3}{(4x^{−1/3})^2}=\dfrac{2^3(x^{2/3})^3}{4^2(x^{−1/3})^2}= \dfrac{8x^2}{16x^{−2/3}} =\dfrac{x^2x^{2/3}}{2}=\dfrac{x^{8/3}}{2}. \nonumber \]

      b. We can simplify as follows:

      \[\dfrac{(x^3y^{−1})^2}{(xy^2)^{−2}}=\dfrac{(x^3)^2(y^{−1})^2}{x^{−2}(y^2)^{−2}}=\dfrac{x^6y^{−2}}{x^{−2}y^{−4}} =x^6x^2y^{−2}y^4=x^8y^2. \nonumber \]

      Exercise \(\PageIndex{2}\)

      Use the laws of exponents to simplify \(\dfrac{6x^{−3}y^2}{12x^{−4}y^5}\).

      Hint

      \(x^a/x^b=x^{a-b}\)

      Answer

      \(x/(2y^3)\)

      The Number e

      A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests \(P\) dollars in a savings account with an annual interest rate \(r\), compounded annually. The amount of money after 1 year is

      \(A(1)=P+rP=P(1+r)\).

      The amount of money after \(2\) years is

      \(A(2)=A(1)+rA(1)=P(1+r)+rP(1+r)=P(1+r)^2\).

      More generally, the amount after \(t\) years is

      \(A(t)=P(1+r)^t\).

      If the money is compounded 2 times per year, the amount of money after half a year is

      \(A\left(\dfrac{1}{2}\right)=P+\left(\dfrac{r}{2}\right)P=P\left(1+\left(\dfrac{r}{2}\right)\right)\).

      The amount of money after \(1\) year is

      \(A(1)=A\left(\dfrac{1}{2}\right)+\left(\dfrac{r}{2}\right)A \left(\dfrac{1}{2}\right)=P\left(1+\dfrac{r}{2}\right)+\dfrac{r}{2}\left(\left(P(1+\dfrac{r}{2}\right)\right)=P\left(1+\dfrac{r}{2}\right)^2.\)

      After \(t\) years, the amount of money in the account is

      \(A(t)=P\left(1+\dfrac{r}{2}\right)^{2t}\).

      More generally, if the money is compounded \(n\) times per year, the amount of money in the account after \(t\) years is given by the function

      \(A(t)=P\left(1+\dfrac{r}{n}\right)^{nt}.\)

      What happens as \(n→∞?\) To answer this question, we let \(m=n/r\) and write

      \(\left(1+\dfrac{r}{n}\right)^{nt}=\left(1+\dfrac{1}{m}\right)^{mrt},\)

      and examine the behavior of \((1+1/m)^m\) as \(m→∞\), using a table of values (Table \(\PageIndex{3}\)).

      Table \(\PageIndex{3}\): Values of \(\left(1+\dfrac{1}{m}\right)^m\) as \(m→∞\)
      \(m\) 10 100 1000 10,000 100,000 1,000,000
      \(\left(1+\dfrac{1}{m}\right)^m\) 2.5937 2.7048 2.71692 2.71815 2.718268 2.718280

      Looking at this table, it appears that \((1+1/m)^m\) is approaching a number between \(2.7\) and \(2.8\) as \(m→∞\). In fact, \((1+1/m)^m\) does approach some number as \(m→∞\). We call this number \(e\). To six decimal places of accuracy,

      \[e≈2.718282. \nonumber \]

      Leonhard Euler

      The letter \(e\) was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although Euler did not discover the number, he showed many important connections between \(e\) and logarithmic functions. We still use the notation \(e\) today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications.

      Returning to our savings account example, we can conclude that if a person puts \(P\) dollars in an account at an annual interest rate \(r\), compounded continuously, then \(A(t)=Pe^{rt}\). This function may be familiar. Since functions involving base \(e\) arise often in applications, we call the function \(f(x)=e^x\) the natural exponential function. Not only is this function interesting because of the definition of the number \(e\), but also, as discussed next, its graph has an important property.

      Since \(e>1\), we know \(f(x) = e^x\) is increasing on \((−∞,∞)\). In Figure \(\PageIndex{3}\), we show a graph of \(f(x)=e^x\) along with a tangent line to the graph of \(f\) at \(x=0\). We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of \(f\) at \(x=a\) is a line that passes through the point \((a,f(a))\) and has the same “slope” as \(f\) at that point . The function \(f(x)=e^x\) is the only exponential function \(b^x\) with tangent line at \(x=0\) that has a slope of \(1.\) As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances.

      1.5: Exponential and Logarithmic Functions (4)
      Example \(\PageIndex{3}\): Compounding Interest

      Suppose \($500\) is invested in an account at an annual interest rate of \(r=5.5\%\), compounded continuously.

      1. Let \(t\) denote the number of years after the initial investment and \(A(t)\) denote the amount of money in the account at time \(t\). Find a formula for \(A(t)\).
      2. Find the amount of money in the account after \(10\) years and after \(20\) years.

      Solution

      a. If \(P\) dollars are invested in an account at an annual interest rate \(r\), compounded continuously, then \(A(t)=Pe^{rt}\). Here \(P=$500\) and \(r=0.055\). Therefore, \(A(t)=500e^{0.055t}\).

      b. After \(10\) years, the amount of money in the account is

      \(A(10)=500e^{0.055⋅10}=500e^{0.55}≈$866.63\).

      After \(20\) years, the amount of money in the account is

      \(A(20)=500e^{0.055⋅20}=500e^{1.1}≈$1,502.08\).

      Exercise \(\PageIndex{3}\)

      If \($750\) is invested in an account at an annual interest rate of \(4\%\), compounded continuously, find a formula for the amount of money in the account after \(t\) years. Find the amount of money after \(30\) years.

      Hint

      \(A(t)=Pe^{rt}\)

      Answer

      \(A(t)=750e^{0.04t}\). After \(30\) years, there will be approximately \($2,490.09\).

      Logarithmic Functions

      Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as the pH scale in chemistry or decibels in sound levels.

      The exponential function \(f(x)=b^x\) is one-to-one, with domain \((−∞,∞)\) and range \((0,∞)\). Therefore, it has an inverse function, called the logarithmic function with base \(b\). For any \(b>0,\, b≠1\), the logarithmic function with base \(b\), denoted \(\log_b\), has domain \((0,∞)\) and range \((−∞,∞)\),and satisfies

      \[\log_b(x)=y \nonumber \]

      if and only if \(b^y=x\).

      For example,

      \[\log_2(8)=3\nonumber \]

      since \(2^3=8\),

      \[\log_{10}\left(\dfrac{1}{100}\right)=−2 \nonumber \]

      since \(10^{−2}=\dfrac{1}{10^2}=\dfrac{1}{100}\),

      \[\log_b(1)=0 \nonumber \]

      since \(b^0=1\) for any base \(b>0\).

      Furthermore, since \(y=\log_b(x)\) and \(y=b^x\) are inverse functions,

      \[\log_b(b^x)=x \nonumber \]

      and

      \[b^{\log_b(x)}=x. \nonumber \]

      The most commonly used logarithmic function is the function \(\log_e\). Since this function uses natural \(e\) as its base, it is called the natural logarithm. Here we use the notation \(\ln (x)\) or \(\ln x\) to mean \(\log_e(x)\). For example,

      \[ \begin{align*} \ln (e) &=\log_e(e)=1 \\[4pt] \ln (e^3) &=\log_e(e^3)=3 \\[4pt] \ln (1) &=\log_e(1)=0. \end{align*}\]

      Since the functions \(f(x)=e^x\) and \(g(x)=\ln (x)\) are inverses of each other,

      \(\ln (e^x)=x\) and \(e^{\ln x}=x\),

      and their graphs are symmetric about the line \(y=x\) (Figure \(\PageIndex{4}\)).

      1.5: Exponential and Logarithmic Functions (5)

      In general, for any base \(b>0\), \(b≠1\), the function \(g(x)=\log_b(x)\) is symmetric about the line \(y=x\) with the function \(f(x)=b^x\). Using this fact and the graphs of the exponential functions, we graph functions \(\log_b\) for several values of \(b>1\) ( Figure \(\PageIndex{5}\)).

      1.5: Exponential and Logarithmic Functions (6)

      Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of logarithms.

      Properties of Logarithms

      If \(a,\,b,\,c>0,\,b≠1\), and \(r\) is any real number, then

      • Product property

      \[\log_b(ac)=\log_b(a)+\log_b(c) \label{productprop} \]

      • Quotient property

      \[\log_b \left(\dfrac{a}{c} \right)=\log_b(a)−\log_b(c) \label{quotientprop} \]

      • Power property

      \[\log_b(a^r)=r\log_b(a) \label{powerprop} \]

      Example \(\PageIndex{4}\): Solving Equations Involving Exponential Functions

      Solve each of the following equations for \(x\).

      1. \(5^x=2\)
      2. \(e^x+6e^{−x}=5\)

      Solution

      a. Applying the natural logarithm function to both sides of the equation, we have

      \(\ln 5^x=\ln 2\).

      Using the power property of logarithms,

      \(x\ln 5=\ln 2.\)

      Therefore,

      \[x= \dfrac{\ln 2}{\ln 5}. \nonumber \]

      b. Multiplying both sides of the equation by \(e^x\),we arrive at the equation

      \(e^{2x}+6=5e^x\).

      Rewriting this equation as

      \(e^{2x}−5e^x+6=0\),

      we can then rewrite it as a quadratic equation in \(e^x\):

      \((e^x)^2−5(e^x)+6=0.\)

      Now we can solve the quadratic equation. Factoring this equation, we obtain

      \((e^x−3)(e^x−2)=0.\)

      Therefore, the solutions satisfy \(e^x=3\) and \(e^x=2\). Taking the natural logarithm of both sides gives us the solutions \(x=\ln 3,\ln 2\).

      Exercise \(\PageIndex{4}\)

      Solve

      \[e^{2x}/(3+e^{2x})=1/2. \nonumber \]

      Hint

      First solve the equation for \(e^{2x}\)

      Answer

      \(x=\dfrac{\ln 3}{2}\).

      Example \(\PageIndex{5}\): Solving Equations Involving Logarithmic Functions

      Solve each of the following equations for \(x\).

      1. \(\ln \left(\dfrac{1}{x}\right)=4\)
      2. \(\log_{10}\sqrt{x}+\log_{10}x=2\)
      3. \(\ln (2x)−3\ln (x^2)=0\)

      Solution

      a. By the definition of the natural logarithm function,

      \(\ln \left(\dfrac{1}{x} \right)=4\)

      • if and only if \(e^4=\dfrac{1}{x}\).

      Therefore, the solution is \(x=1/e^4\).

      b. Using the product (Equation \ref{productprop}) and power (Equation \ref{powerprop}) properties of logarithmic functions, rewrite the left-hand side of the equation as

      \[\begin{align*} \log_{10}\sqrt{x} + \log_{10}x &= \log_{10} x \sqrt{x} \\[4pt] &= \log_{10}x^{3/2} \\[4pt] &= \dfrac{3}{2}\log_{10}x. \end{align*}\]

      Therefore, the equation can be rewritten as

      \(\dfrac{3}{2}\log_{10}x=2\)

      or

      \(\log_{10}x=\dfrac{4}{3}\).

      The solution is \(x=10^{4/3}=10\sqrt[3]{10}\).

      c. Using the power property (Equation \ref{powerprop}) of logarithmic functions, we can rewrite the equation as \(\ln (2x)−\ln (x^6)=0\).

      Using the quotient property (Equation \ref{quotientprop}), this becomes

      \(\ln \left(\dfrac{2}{x^5}\right)=0\)

      Therefore, \(2/x^5=1\), which implies \(x=\sqrt[5]{2}\). We should then check for any extraneous solutions.

      Exercise \(\PageIndex{5}\)

      Solve \(\ln (x^3)−4\ln (x)=1\).

      Hint

      First use the power property, then use the product property of logarithms.

      Answer

      \(x=\dfrac{1}{e}\)

      When evaluating a logarithmic function with a calculator, you may have noticed that the only options are \(\log_{10}\) or \(\log\), called the common logarithm, or \(\ln\), which is the natural logarithm. However, exponential functions and logarithm functions can be expressed in terms of any desired base \(b\). If you need to use a calculator to evaluate an expression with a different base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or logarithmic function in terms of the natural exponential and natural logarithmic functions.

      Rule: Change-of-Base Formulas

      Let \(a>0,\,b>0\), and \(a≠1,\,b≠1\).

      1. \(a^x=b^{x \log_ba}\) for any real number \(x\).

      If \(b=e\), this equation reduces to \(a^x=e^{x \log_ea}=e^{x \ln a}\).

      2. \(\log_ax=\dfrac{\log_bx}{\log_ba}\) for any real number \(x>0\).

      If \(b=e\), this equation reduces to \(\log_ax=\dfrac{\ln x}{\ln a}\).

      Proof

      For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that for any base \(b>0,\, b≠1\), \(\log_b(a^x)=x \log_ba\). Therefore,

      \(b^{\log_b(a^x)}\)=\(b^{x \log_ba}\).

      In addition, we know that \(b^x\) and \(\log_b(x)\) are inverse functions. Therefore,

      \(b^{\log_b(a^x)}=a^x\).

      Combining these last two equalities, we conclude that \(a^x=b^{x \log_ba}\).

      To prove the second property, we show that

      \((\log_ba)⋅(\log_ax)=\log_bx.\)

      Let \(u=\log_ba,v=\log_ax\), and \(w=\log_bx\). We will show that \(u⋅v=w\). By the definition of logarithmic functions, we know that \(b^u=a,\, a^v=x\), and \(b^w=x\). From the previous equations, we see that

      \(b^{uv}=(b^u)^v=a^v=x=b^w.\)

      Therefore, \(b^{uv}=b^w\). Since exponential functions are one-to-one, we can conclude that \(u⋅v=w\).

      \(\square\)

      Example \(\PageIndex{6}\): Changing Bases

      Use a calculating utility to evaluate \(\log_37\) with the change-of-base formula presented earlier.

      Solution

      Use the second equation with \(a=3\) and \(b=e\): \(\log_37=\dfrac{\ln 7}{\ln 3}≈1.77124\).

      Exercise \(\PageIndex{6}\)

      Use the change-of-base formula and a calculating utility to evaluate \(\log_46\).

      Hint

      Use the change of base to rewrite this expression in terms of expressions involving the natural logarithm function.

      Answer

      \(\log_46 = \dfrac{\ln 6}{\ln 4} \approx 1.29248\)

      Example \(\PageIndex{7}\): The Richter Scale for Earthquakes

      In 1935, Charles Richter developed a scale (now known as the Richter scale) to measure the magnitude of an earthquake. The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake with magnitude \(R_1\) on the Richter scale and a second earthquake with magnitude \(R_2\) on the Richter scale. Suppose \(R_1>R_2\), which means the earthquake of magnitude \(R_1\) is stronger, but how much stronger is it than the other earthquake?

      1.5: Exponential and Logarithmic Functions (7)

      A way of measuring the intensity of an earthquake is by using a seismograph to measure the amplitude of the earthquake waves. If \(A_1\) is the amplitude measured for the first earthquake and \(A_2\) is the amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy the following equation:

      \(R_1−R_2=\log_{10}\left(\dfrac{A1}{A2}\right)\).

      Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter scale. Then,

      \(8−7=\log_{10}\left(\dfrac{A1}{A2}\right)\).

      Therefore,

      \(\log_{10}\left(\dfrac{A1}{A2}\right)=1\),

      which implies \(A_1/A_2=10\) or \(A_1=10A_2\). Since \(A_1\) is 10 times the size of \(A_2\), we say that the first earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation

      \(\log_{10}\left(\dfrac{A1}{A2}\right)=8−6=2\).

      Therefore, \(A_1=100A_2\).That is, the first earthquake is 100 times more intense than the second earthquake.

      How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in 2011 with the magnitude 7.3 earthquake in Haiti in 2010?

      Solution

      To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:

      \(9−7.3=\log_{10}\left(\dfrac{A1}{A2}\right)\).

      Therefore, \(A_1/A_2=10^{1.7}\), and we conclude that the earthquake in Japan was approximately 50 times more intense than the earthquake in Haiti.

      Exercise \(\PageIndex{7}\)

      Compare the relative severity of a magnitude \(8.4\) earthquake with a magnitude \(7.4\) earthquake.

      Hint

      \(R_1−R_2=\log_{10}(A1/A2)\).

      Answer

      The magnitude \(8.4\) earthquake is roughly \(10\) times as severe as the magnitude \(7.4\) earthquake.

      Hyperbolic Functions

      The hyperbolic functions are defined in terms of certain combinations of \(e^x\) and \(e^{−x}\). These functions arise naturally in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes. Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary (Figure \(\PageIndex{7}\)). If we introduce a coordinate system so that the low point of the chain lies along the \(y\)-axis, we can describe the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.

      1.5: Exponential and Logarithmic Functions (8)
      Definitions: hyperbolic functions

      Hyperbolic cosine

      \(\cosh x=\dfrac{e^x+e^{−x}}{2}\)

      Hyperbolic sine

      \(\sinh x=\dfrac{e^x−e^{−x}}{2}\)

      Hyperbolic tangent

      \(\tanh x=\dfrac{\sinh x}{\cosh x}=\dfrac{e^x−e^{−x}}{e^x+e^{−x}}\)

      Hyperbolic cosecant

      \(\operatorname{csch}x=\dfrac{1}{\sinh x}=\dfrac{2}{e^x−e^{−x}}\)

      Hyperbolic secant

      \(\operatorname{sech}x=\dfrac{1}{\cosh x}=\dfrac{2}{e^x+e^{−x}}\)

      Hyperbolic cotangent

      \(\coth x=\dfrac{\cosh x}{\sinh x}=\dfrac{e^x+e^{−x}}{e^x−e^{−x}}\)

      The name \(\cosh\) rhymes with “gosh,” whereas the name \(\sinh\) is pronounced “cinch.” \(\operatorname{Tanh}, \,\operatorname{sech}, \, \operatorname{csch},\) and \(\coth\) are pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.

      Using the definition of \(\cosh(x)\) and principles of physics, it can be shown that the height of a hanging chain, such as the one in Figure \(\PageIndex{8}\), can be described by the function \(h(x)=a\cosh(x/a) + c\) for certain constants \(a\) and \(c\).

      But why are these functions called hyperbolic functions? To answer this question, consider the quantity \(\cosh^2 t − \sinh^2 t\). Using the definition of \(\cosh\) and \(\sinh\), we see that

      \[\cosh^2 t − \sinh^2 t=\dfrac{e^{2t}+2+e^{−2t}}{4}−\dfrac{e^{2t}−2+e^{−2t}}{4}=1. \nonumber \]

      This identity is the analog of the trigonometric identity \(\cos^2 t + \sin^2 t=1\). Here, given a value \(t\), the point \((x,y)=(\cosh t,\,\sinh t)\) lies on the unit hyperbola \(x^2−y^2=1\) (Figure \(\PageIndex{8}\)).

      1.5: Exponential and Logarithmic Functions (9)

      Graphs of Hyperbolic Functions

      To graph \(\cosh x\) and \(\sinh x\), we make use of the fact that both functions approach \((1/2)e^x\) as \(x→∞\), since \(e^{−x}→0\) as \(x→∞\). As \(x→−∞,\cosh x\) approaches \(1/2e^{−x}\), whereas \(\sinh x\) approaches \(−1/2e^{−x}\). Therefore, using the graphs of \(1/2e^x,1/2e^{−x}\), and \(−1/2e^{−x}\) as guides, we graph \(\cosh x\) and \(\sinh x\). To graph \(\tanh x\), we use the fact that \(\tanh(0)=0\), \(−1<\tanh(x)<1\) for all \(x\), \(\tanh x→1\) as \(x→∞\), and \(\tanh x→−1\) as \(x→−∞\). The graphs of the other three hyperbolic functions can be sketched using the graphs of \(\cosh x\), \(\sinh x\), and \(\tanh x\) (Figure \(\PageIndex{9}\)).

      1.5: Exponential and Logarithmic Functions (10)

      Identities Involving Hyperbolic Functions

      The identity \(\cosh^2 t−\sinh^2 t = 1\), shown in Figure \(\PageIndex{8}\), is one of several identities involving the hyperbolic functions, some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperbolic cosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.

      Identities Involving Hyperbolic Functions
      1. \(\cosh(−x)=\cosh x\)
      2. \(\sinh(−x)=−\sinh x\)
      3. \(\cosh x+\sinh x=e^x\)
      4. \(\cosh x−\sinh x=e^{−x}\)
      5. \(\cosh^2 x−\sinh^2 x=1\)
      6. \(1−\tanh^2 x=\operatorname{sech}^2 x\)
      7. \(\coth^2 x −1=\operatorname{csch}^2 x\)
      8. \(\sinh(x±y)=\sinh x \cosh y ± \cosh x \sinh y\)
      9. \(\cosh(x±y)=\cosh x \cosh y ± \sinh x \sinh y\)
        Example \(\PageIndex{8}\): Evaluating Hyperbolic Functions
        1. Simplify \(\sinh(5\ln x)\).
        2. If \(\sinh x=3/4\), find the values of the remaining five hyperbolic functions.

        Solution:

        a. Using the definition of the \(\sinh\) function, we write

        \(\sinh(5\ln x)=\dfrac{e^{5\ln x}−e^{−5\ln x}}{2}=\dfrac{e^{\ln (x^5)}−e^{\ln (x^{−5})}}{2}=\dfrac{x^5−x^{−5}}{2}.\)

        b. Using the identity \(\cosh^2 x − \sinh^2 x=1\),we see that

        \(\cosh^2 x=1+\left(\frac{3}{4}\right)^2=\dfrac{25}{16}.\)

        Since \(\cosh x≥1\) for all \(x\), we must have \(\cosh x=5/4\). Then, using the definitions for the other hyperbolic functions, we conclude that \(\tanh x=3/5,\operatorname{csch}x=4/3,\operatorname{sech}x=4/5\), and \(\coth x=5/3\).

        Exercise \(\PageIndex{8}\)

        Simplify \(\cosh(2\ln x)\).

        Hint

        Use the definition of the \(\cosh\) function and the power property of logarithm functions.

        Answer

        \((x^2+x^{−2})/2\)

        Inverse Hyperbolic Functions

        From the graphs of the hyperbolic functions, we see that all of them are one-to-one except \(\cosh x\) and \(\operatorname{sech}x\). If we restrict the domains of these two functions to the interval \([0,∞),\) then all the hyperbolic functions are one-to-one, and we can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the inverse hyperbolic functions involve logarithmic functions.

        Definitions: Inverse Hyperbolic Functions

        \[\begin{align*} &\sinh^{−1}x =\operatorname{arcsinh}x=\ln \left(x+\sqrt{x^2+1}\right) & & \cosh^{−1}x =\operatorname{arccosh}x=\ln \left(x+\sqrt{x^2−1}\right)\\[4pt]
        &\tanh^{−1}x=\operatorname{arctanh}x=\dfrac{1}{2}\ln \left(\dfrac{1+x}{1−x}\right) & & \coth^{−1}x =\operatorname{arccot}x=\frac{1}{2}\ln \left(\dfrac{x+1}{x−1}\right)\\[4pt]
        &\operatorname{sech}^{−1}x=\operatorname{arcsech}x=\ln \left(\dfrac{1+\sqrt{1−x^2}}{x}\right) & & \operatorname{csch}^{−1}x=\operatorname{arccsch}x=\ln \left(\dfrac{1}{x}+\dfrac{\sqrt{1+x^2}}{|x|}\right) \end{align*}\]

        Let’s look at how to derive the first equation. The others follow similarly. Suppose \(y=\sinh^{−1}x\). Then, \(x=\sinh y\) and, by the definition of the hyperbolic sine function, \(x=\dfrac{e^y−e^{−y}}{2}\). Therefore,

        \(e^y−2x−e^{−y}=0.\)

        Multiplying this equation by \(e^y\), we obtain

        \(e^{2y}−2xe^y−1=0\).

        This can be solved like a quadratic equation, with the solution

        \(e^y=\dfrac{2x±\sqrt{4x^2+4}}{2}=x±\sqrt{x^2+1}\).

        Since \(e^y>0\),the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the equation, we conclude that

        \(y=\ln (x+\sqrt{x^2+1}).\)

        Example \(\PageIndex{9}\): Evaluating Inverse Hyperbolic Functions

        Evaluate each of the following expressions.

        \(\sinh^{−1}(2)\)

        \(\tanh^{−1}(1/4)\)

        Solution:

        \[\sinh^{−1}(2)=\ln (2+\sqrt{2^2+1})=\ln (2+\sqrt{5})≈1.4436\nonumber \]

        \[\tanh^{−1}(1/4)=\frac{1}{2}\ln \left(\dfrac{1+1/4}{1−1/4}\right)=\frac{1}{2}\ln \left(\dfrac{5/4}{3/4}\right)=\frac{1}{2}\ln \left(\dfrac{5}{3}\right)≈0.2554\nonumber \]

        Exercise \(\PageIndex{9}\)

        Evaluate \(\tanh^{−1}(1/2)\).

        Hint

        Use the definition of \(\tanh^{−1}x\) and simplify.

        Answer

        \(\dfrac{1}{2}\ln (3)≈0.5493\).

        Key Concepts

        • The exponential function \(y=b^x\) is increasing if \(b>1\) and decreasing if \(0<b<1\). Its domain is \((−∞,∞)\) and its range is \((0,∞)\).
        • The logarithmic function \(y=\log_b(x)\) is the inverse of \(y=b^x\). Its domain is \((0,∞)\) and its range is \((−∞,∞).\)
        • The natural exponential function is \(y=e^x\) and the natural logarithmic function is \(y=\ln x=\log_ex.\)
        • Given an exponential function or logarithmic function in base \(a\), we can make a change of base to convert this function to any base \(b>0\), \(b≠1.\) We typically convert to base \(e\).
        • The hyperbolic functions involve combinations of the exponential functions \(e^x\) and \(e^{−x}.\) As a result, the inverse hyperbolic functions involve the natural logarithm.

        Glossary

        base
        the number \(b\) in the exponential function \(f(x)=b^x\) and the logarithmic function \(f(x)=\log_bx\)
        exponent
        the value \(x\) in the expression \(b^x\)
        hyperbolic functions
        the functions denoted \(\sinh,\,\cosh,\,\operatorname{tanh},\,\operatorname{csch},\,\operatorname{sech},\) and \(\coth\), which involve certain combinations of \(e^x\) and \(e^{−x}\)
        inverse hyperbolic functions
        the inverses of the hyperbolic functions where \(\cosh\) and \( \operatorname{sech}\) are restricted to the domain \([0,∞)\);each of these functions can be expressed in terms of a composition of the natural logarithm function and an algebraic function
        natural exponential function
        the function \(f(x)=e^x\)
        natural logarithm
        the function \(\ln x=\log_ex\)
        number e
        as \(m\) gets larger, the quantity \((1+(1/m)^m\) gets closer to some real number; we define that real number to be \(e;\) the value of \(e\) is approximately \(2.718282\)
        1.5: Exponential and Logarithmic Functions (2024)

        FAQs

        Are exponential functions always 1 to 1? ›

        Apart from such normal cases, exponential functions are often not one-one.

        How can you solve exponential and logarithmic equations? ›

        Step 1: Isolate the exponential expression. Step 2: Take the logarithm of both sides. In this case, we will take the common logarithm of both sides so that we can approximate our result on a calculator. Step 3: Apply the power rule for logarithms and then solve.

        How do you tell if a function is exponential or logarithmic? ›

        An exponential function has the form ax, where a is a constant; examples are 2x, 10x, ex. The logarithmic functions are the inverses of the exponential functions, that is, functions that "undo'' the exponential functions, just as, for example, the cube root function "undoes'' the cube function: 3√23=2.

        Can a function be 1 to 1 but not onto? ›

        in an 'onto' function, every x -value is mapped to a y− value. in a one-to-one function, every y -value is mapped to at most one x - value. this means that in a one-to-one function, not every x -value in the domain must be mapped on the graph. it only means that no y -value can be mapped twice.

        Can an exponential function be less than 1? ›

        In an exponential function, the base can be any number greater than 0, except 1. If the base is greater than 1, the function calculates exponential growth. If it is greater than 0 but less than 1, the function calculates exponential decay. The base serves as a starting point for calculating the growth or decline.

        What are the rules for exponential and logarithmic functions? ›

        We saw earlier that an exponential function is any function of the form f(x)=bx, where b>0 and b≠1. A logarithmic function is any function of the form g(x)=logb(x), where b>0 and b≠1. It is no coincidence that both forms have the same restrictions on b because they are inverses of each other.

        What are the rules for solving exponential functions? ›

        The first law states that to multiply two exponential functions with the same base, we simply add the exponents. The second law states that to divide two exponential functions with the same base, we subtract the exponents. The third law states that in order to raise a power to a new power, we multiply the exponents.

        What is the difference between logarithmic and exponential equations? ›

        An exponential equation is an equation in which the variable appears in an exponent. A logarithmic equation is an equation that involves the logarithm of an expression containing a variable.

        Can an exponential function be zero? ›

        The function ex considered as a function of Real numbers has domain (−∞,∞) and range (0,∞) . So it can only take strictly positive values. When we consider ex as a function of Complex numbers, then we find it has domain C and range C\{0} . That is 0 is the only value that ex cannot take.

        How to convert exponential function to logarithmic function? ›

        The process of converting from exponential to log form is a simple process. The exponential form ax=N a x = N is converted to logarithmic form logaN=x l o g a N = x , and this simple formula is helpful to convert exponential to log form.

        How to solve logarithms easily? ›

        The questions of logarithm could be solved based on the properties, given below:
        1. Product rule: logb MN = logb M + logb N.
        2. Quotient rule: logb M/N = logb M – logb N.
        3. Power rule: logb Mp = P logb M.
        4. Zero Exponent Rule: loga 1 = 0.
        5. Change of Base Rule: logb (x) = ln x / ln b or logb (x) = log10 x / log10 b.

        Are logarithms difficult? ›

        While working on the problem of logarithms, students are often constrained by the many logarithmic properties and difficult for them to memorize. The constraint makes it difficult for students to understand logarithms, due to over-focusing on the theorems or properties they need to memorize.

        How do you solve logarithmic equations for dummies? ›

        Turn the variable inside the log into an exponential equation (which is all about the base, of course). For example, to solve log3 x = –4, change it to the exponential equation 34 = x, or 1/81 = x. Keep in mind that because logs don't have negative bases, you throw the negative one out the window and say x = 4 only.

        How to know if an exponential function is one-to-one? ›

        Recall that an exponential function f : R → (0,с) has as its domain the set R and has as its target the set (0,с). We see from the graph of f(x) = ax, if either a > 1 or 0 <a< 1, that f(x) is one-to-one and onto. Remember that to check if f(x) is one-to-one, we can use the horizontal line test (which f(x) passes).

        Is something to the power of 1 always 1? ›

        But it's actually pretty simple! One to the power of ANY NUMBER will always be one because you just keep on doing 1 x 1 x 1 x 1 etc.

        Do exponential functions always go through 0 1? ›

        Expert-Verified Answer

        Exponential functions always pass through 0 1 because they always rise to the left and approach 0 to the right.

        What is the rule of exponential functions? ›

        The first law states that to multiply two exponential functions with the same base, we simply add the exponents. The second law states that to divide two exponential functions with the same base, we subtract the exponents. The third law states that in order to raise a power to a new power, we multiply the exponents.

        References

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