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Chapter 3: Problem 69
Solve the exponential equation algebraically. Approximate the result to three decimal places. $$\left(1+\frac{0.10}{12}\right)^{12 t}=2$$
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Use the acidity model given by \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right],\) where acidity \((\mathrm{pH})\) is a measure of the hydrogen ion concentration \(\left[\mathrm{H}^{+}\right]\) (measured in moles of hydrogen per liter) of a solution. Compute \(\left[\mathrm{H}^{+}\right]\) for a solution in which \(\mathrm{pH}=5.8\).
Determine the principal \(P\) that must be invested at rate \(r\), compounded monthly, so that $$\$ 500,000$$ will be available for retirement in \(t\) years. $$r=5 \%, t=10$$
The demand equation for a limited edition coin set is \(p=1000\left(1-\frac{5}{5+e^{-0.001 x}}\right)\) Find the demand \(x\) for a price of (a) \(p=\$ 139.50\) and (b) \(p=\$ 99.99\).
(a) solve for \(P\) and (b) solve for \(t\). $$A=P e^{r t}$$
At 8: 30 A.M., a coroner was called 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 is derived 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.
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