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Chapter 3: Problem 56
Use the One-to-One Property to solve the equation for \(x\). $$e^{2 x-1}=e^{4}$$
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Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log 8 x-\log (1+\sqrt{x})=2$$
A $$\$ 120,000$$ home mortgage for 30 years at \(7 \frac{1}{2} \%\) has a monthly payment of $$\$ 839.06 .$$ Part of the monthly payment is paid toward the interest charge on the unpaid balance, and the remainder of the payment is used to reduce the principal. The amount that is paid toward the interest is \(u=M-\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) and the amount that is paid toward the reduction of the principal is \(v=\left(M-\frac{P r}{12}\right)\left(1+\frac{r}{12}\right)^{12 t}\) In these formulas, \(P\) is the size of the mortgage, \(r\) is the interest rate, \(M\) is the monthly payment, and \(t\) is the time (in years). (a) Use a graphing utility to graph each function in the same viewing window. (The viewing window should show all 30 years of mortgage payments.) (b) In the early years of the mortgage, is the larger part of the monthly payment paid toward the interest or the principal? Approximate the time when the monthly payment is evenly divided between interest and principal reduction. (c) Repeat parts (a) and (b) for a repayment period of 20 years \((M=\$ 966.71) .\) What can you conclude?
Carbon 14 dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If this is true, the amount of \({ }^{14} \mathrm{C}\) absorbed by a tree that grew several centuries ago should be the same as the amount of \({ }^{14} \mathrm{C}\) absorbed by a tree growing today. A piece of ancient charcoal contains only \(15 \%\) as much radioactive carbon as a piece of modern charcoal. How long ago was the tree burned to make the ancient charcoal if the half-life of \({ }^{14} \mathrm{C}\) is 5715 years?
Determine the time necessary for $$\$ 1000$$to double if it is invested at interest rate \(r\) compounded (a) annually, (b) monthly, (c) daily, and (d) continuously. $$r=6.5 \%$$
Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$3-\ln x=0$$
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