Chapter 3: Problem 16
Write each logarithmic equation in its equivalent exponential form. $$\ln 4=y$$
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Chapter 3: Problem 16
Write each logarithmic equation in its equivalent exponential form. $$\ln 4=y$$
These are the key concepts you need to understand to accurately answer the question.
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$$\text { Graph the function } f(x)=\left\\{\begin{array}{ll}-\ln (-x) & x<0 \\\\-\ln (x) & x>0\end{array}\right.$$
Solve the equation \(\ln 3 x=\ln \left(x^{2}+1\right) .\) Using a graphing calculator, plot the graphs \(y=\ln (3 x)\) and \(y=\ln \left(x^{2}+1\right)\) in the same viewing rectangle. Zoom in on the point where the graphs intersect. Does this agree with your solution?
Use a graphing calculator to plot \(y=\log x\) and \(y=\frac{\ln x}{\ln 10}\) Are they the same graph?
Amy has a credit card debt in the amount of \(\$ 12,000 .\) The annual interest is \(18 \% .\) Her time \(t\) to pay off the loan is given by $$t=-\frac{\ln \left[1-\frac{12,000(0.18)}{n R}\right]}{n \ln \left(1+\frac{0.18}{n}\right)}$$ where \(n\) is the number of payment periods per year and \(R\) is the periodic payment. a. Use a graphing utility to graph $$t_{1}=-\frac{\ln \left[1-\frac{12,000(0.18)}{12 x}\right]}{12 \ln \left(1+\frac{0.18}{12}\right)} \text { as } Y_{1} \text { and }$$ $$t_{2}=-\frac{\ln \left[1-\frac{12,000(0.18)}{26 x}\right]}{26 \ln \left(1+\frac{0.18}{26}\right)} \text { as } Y_{2}$$ Explain the difference in the two graphs. b. Use the \([\text { TRACE }]\) key to estimate the number of years that it will take Amy to pay off her credit card if she can afford a monthly payment of \(\$ 300 .\) c. If she can make a biweekly payment of \(\$ 150,\) estimate the number of years that it will take her to pay off the credit card. d. If Amy adds \(\$ 100\) more to her monthly or \(\$ 50\) more to her biweekly payment, estimate the number of years that it will take her to pay off the credit card.
Determine whether each statement is true or false. The division of two logarithms with the same base is equal to the logarithm of the subtraction.
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