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Chapter 3: Problem 25
Solve the exponential equation algebraically. Approximate the result to three decimal places. $$2^{3-x}=565$$
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Forensics At 8: 30 A.M., a coroner went to the home of a person who had died during the night. In order to estimate the time of death, the coroner took the person's temperature twice. At 9: 00 A.M. the temperature was \(85.7^{\circ} \mathrm{F},\) and at 11: 00 A.M. the temperature was \(82.8^{\circ} \mathrm{F}\). From these two temperatures, the coroner was able to determine that the time elapsed since death and the body temperature were related by the formula $$t=-10 \ln \frac{T-70}{98.6-70}$$ where \(t\) is the time in hours elapsed since the person died and \(T\) is the temperature (in degrees Fahrenheit) of the person's body. (This formula comes from a general cooling principle called Newton's Law of Cooling. It uses the assumptions that the person had a normal body temperature of \(98.6^{\circ} \mathrm{F}\) at death and that the room temperature was a constant \(70^{\circ} \mathrm{F}\).) Use the formula to estimate the time of death of the person.
The values \(y\) (in billions of dollars) of U.S. currency in circulation in the years 2000 through 2010 can be modeled by \(y=-611+507\) ln \(t, 10 \leq t \leq 20\) where \(t\) represents the year, with \(t=10\) corresponding to 2000. During which year did the value of U.S. currency in circulation exceed \(\$ 690\) billion? (Source: Board of Governors of the Federal Reserve System )
Writing a Natural Logarithmic Equation In Exercises \(53-56,\) write the exponential equation in logarithmic form. $$e^{1 / 2}=1.6487 \ldots$$
Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$6 \log _{3}(0.5 x)=11$$
Population The time \(t\) (in years) for the world population to double when it is increasing at a continuous rate of \(r\) is given by \(t=(\ln 2) / r\) (a) Complete the table and interpret your results. \begin{tabular}{|l|l|l|l|l|l|l|}\hline\(r\) & 0.005 & 0.010 & 0.015 & 0.020 & 0.025 & 0.030 \\\\\hline\(t\) & & & && &..\begin{array}{|l|l|l|l|l|l|l|} \hline r & 0.005 & 0.010 & 0.015 & 0.020 & 0.025 & 0.030 \\ \hline t & & & & & & \\ \hline \end{array}
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