Chapter 4

Evaluate the expression without using a calculator.\(\ln e^{-4}\)

Endangered Species A conservation organization releases 100 animals of an endangered species into a game preserve. The organization believes that the preserve has a carrying capacity of 1000 animals and that the growth of the herd will be modeled by the logistic curve \(p=\frac{1000}{1+9 e^{-k t}}, \quad t \geq 0\) where \(p\) is the number of animals and \(t\) is the time (in years). The herd size is 134 after 2 years. Find \(k\). Then find the population after 5 years.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(8\left(3^{6-x}\right)=40\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} \frac{3}{5}\)

Evaluate the expression without using a calculator.\(\ln \frac{1}{e^{3}}\)

Aged Population The table shows the projected U.S. populations \(P\) (in thousands) of people who are 85 years old or older for several years from 2010 to \(2050 . \quad\) (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & 85 \text { years and older } \\ \hline 2010 & 6123 \\ \hline 2015 & 6822 \\ \hline 2020 & 7269 \\ \hline 2025 & 8011 \\ \hline 2030 & 9603 \\ \hline 2035 & 12,430 \\ \hline 2040 & 15,409 \\ \hline 2045 & 18,498 \\ \hline 2050 & 20,861 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=10\) corresponding to 2010 . (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use a graphing utility to graph the exponential model in base \(e\). (d) Use the exponential model in base \(e\) to estimate the populations of people who are 85 years old or older in 2022 and in 2042 .

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{3 x}=12\)

Evaluate the expression without using a calculator.\(\log _{a} a^{5}\)

Super Bowl Ad Cost The table shows the costs \(C\) (in millions of dollars) of a 30 -second TV ad during the Super Bowl for several years from 1987 to \(2006 .\) (Source: TNS Media Intelligence)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Cost } \\ \hline 1987 & 0.6 \\ \hline 1992 & 0.9 \\ \hline 1997 & 1.2 \\ \hline 2002 & 2.2 \\ \hline 2006 & 2.5 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=7\) corresponding to \(1987 .\) (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use a graphing utility to graph the exponential model in base \(e\). (d) Use the exponential model in base \(e\) to predict the costs of a 30 -second ad during the Super Bowl in 2009 and in 2010 .

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}=50\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} 81\)

Evaluate the expression without using a calculator.\(\log _{a} 1\)

Super Bowl Ad Revenue The table shows Super Bowl TV ad revenues \(R\) (in millions of dollars) for several years from 1987 to 2006. (Source: TNS Media Intelligence)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Revenue } \\ \hline 1987 & 31.5 \\ \hline 1992 & 48.2 \\ \hline 1997 & 72.2 \\ \hline 2002 & 134.2 \\ \hline 2006 & 162.5 \\ \hline \end{array} $$(a) Use a spreadsheet software program to create a scatter plot of the data. Let \(t\) represent the year, with \(t=7\) corresponding to 1987 . (b) Use the regression feature of a spreadsheet software program to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use a spreadsheet software program to graph the exponential model in base \(e\). (d) Use the exponential model in base \(e\) to predict the Super Bowl ad revenues in 2009 and in 2010 .

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(500 e^{-x}=300\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} \sqrt{2}\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\log _{10} 345\)

Domestic Demand The domestic demands \(D\) (in thousands of barrels) for refined oil products in the United States from 1995 to 2005 are shown in the table. (Source: U.S. Energy Information Administration)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 1995 & 6,469,625 \\ \hline 1996 & 6,701,094 \\ \hline 1997 & 6,796,300 \\ \hline 1998 & 6,904,705 \\ \hline 1999 & 7,124,435 \\ \hline 2000 & 7,210,566 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Demand } \\ \hline 2001 & 7,171,885 \\ \hline 2002 & 7,212,765 \\ \hline 2003 & 7,312,410 \\ \hline 2004 & 7,587,546 \\ \hline 2005 & 7,539,440 \\ \hline \end{array} $$(a) Use a spreadsheet software program to create a scatter plot of the data. Let \(t\) represent the year, with \(t=5\) corresponding to 1995 . (b) Use the regression feature of a spreadsheet software program to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use the regression feature of a spreadsheet software program to find a logarithmic model \((y=a+b \ln x)\) for the data. (d) Use a spreadsheet software program to graph the exponential model in base \(e\) and the logarithmic model with the scatter plot. (e) Use both models to predict domestic demands in 2008 , 2009, and \(2010 .\) Do both models give reasonable predictions? Explain.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(1000 e^{-4 x}=75\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} \sqrt{5}\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\log _{10} 163\)

Population The populations \(P\) of the United States (in thousands) from 1990 to 2005 are shown in the table. (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1990 & 250,132 \\ \hline 1991 & 253,493 \\ \hline 1992 & 256,894 \\ \hline 1993 & 260,255 \\ \hline 1994 & 263,436 \\ \hline 1995 & 266,557 \\ \hline 1996 & 269,667 \\ \hline 1997 & 272,912 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1998 & 276,115 \\ \hline 1999 & 279,295 \\ \hline 2000 & 282,403 \\ \hline 2001 & 285,335 \\ \hline 2002 & 288,216 \\ \hline 2003 & 291,089 \\ \hline 2004 & 293,908 \\ \hline 2005 & 296,639 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=0\) corresponding to 1990 . (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Use the regression feature of a graphing utility to find a linear model and a quadratic model for the data. (d) Use a graphing utility to graph the exponential model in base \(e\) and the models in part (c) with the scatter plot. (e) Use each model to predict the populations in 2008 , 2009 , and 2010 . Do all models give reasonable predictions? Explain.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(7-2 e^{x}=6\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} 40\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\log _{10} \frac{4}{5}\)

Population The population \(P\) of the United States officially reached 300 million at about 7:46 A.M. E.S.T. on Tuesday, October 17,2006 . The table shows the U.S. populations (in millions) since 1900. (Source: U.S. Census Bureau)$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1900 & 76 \\ \hline 1910 & 92 \\ \hline 1920 & 106 \\ \hline 1930 & 123 \\ \hline 1940 & 132 \\ \hline 1950 & 151 \\ \hline \end{array} $$$$ \begin{array}{|c|c|} \hline \text { Year } & \text { Population } \\ \hline 1960 & 179 \\ \hline 1970 & 203 \\ \hline 1980 & 227 \\ \hline 1990 & 250 \\ \hline 2000 & 282 \\ \hline 2006 & 300 \\ \hline \end{array} $$(a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=0\) corresponding to 1900 . (b) Use the regression feature of a graphing utility to find an exponential model for the data. Use the Inverse Property \(b=e^{\ln b}\) to rewrite the model as an exponential model in base \(e\). (c) Graph the exponential model in base \(e\) with the scatter plot of the data. What appears to be happening to the relationship between the data points and the regression curve at \(t=100\) and \(t=106 ?\) (d) Use the regression feature of a graphing utility to find a logistic growth model for the data. Graph each model using the window settings shown below. Which model do you think will give more accurate predictions of the population well beyond \(2006 ?\)

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(-14+3 e^{x}=11\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} 45\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\log _{10} \frac{3}{4}\)

Compound Interest A bank offers two types of interest-bearing accounts. The first account pays \(6 \%\) interest compounded monthly. The second account pays \(5 \%\) interest compounded continuously. Which account earns more money? Why?

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(6\left(2^{3 x-1}\right)-7=9\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b}(2 b)^{-2}\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\log _{10} \sqrt{8}\)

MAKE A DECISION: CASH SETTLEMENT You invest a cash settlement of \(\$ 10,000\) for 5 years. You have a choice between an account that pays \(6.25 \%\) interest compounded monthly with a monthly online access fee of \(\$ 5\) and an account that pays \(5.25 \%\) interest compounded continuously with free online access. Which account should you choose? Explain your reasoning.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(8\left(4^{6-2 x}\right)+13=41\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b}\left(3 b^{2}\right)\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\log _{10} \sqrt{3}\)

MAKE A DECISION: SALES COMMISSION You invest a sales commission of \(\$ 12,000\) for 6 years. You have a choice between an account that pays \(4.85 \%\) interest compounded monthly with a monthly online access fee of \(\$ 3\) and an account that pays \(4.25 \%\) interest compounded continuously with free online access. Which account should you choose? Explain your reasoning.

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-8 e^{x}+12=0\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} \sqrt[3]{4 b}\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\ln 7\)

(a) \(I=10^{-3}\) watt per square meter (loud car horn) (b) \(I \approx 10^{0}\) watt per square meter (threshold of pain)

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-5 e^{x}+6=0\)

Approximate the logarithm using the properties of logarithms, given \(\log _{b} 2 \approx 0.3562\), \(\log _{b} 3 \approx 0.5646\), and \(\log _{b} 5 \approx 0.8271$$\log _{b} \sqrt[3]{3 b}\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(2 \ln 9\)

Compute \(\left[\mathrm{H}^{+}\right]\) for a solution for which \(\mathrm{pH}=5.8\)

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-3 e^{x}-4=0\)

Find the exact value of the logarithmic expression without using a calculator.\(\log _{4} \sqrt[3]{4}\)

Use a calculator to evaluate the logarithm. Round your result to three decimal places.\(\ln 18.42\)

Compute \(\left[\mathrm{H}^{+}\right]\) for a solution for which \(\mathrm{pH}=7.3\).

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-9 e^{x}-36=0\)