Chapter 3: Problem 5
A logistic growth model has the form ________.
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Chapter 3: Problem 5
A logistic growth model has the form ________.
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Solve the exponential equation algebraically. Approximate the result to three decimal places. $$e^{3 x}=12$$
An exponential growth model has the form ________ and an exponential decay model has the form ________.
The model $$t=16.625 \ln \left(\frac{x}{x-750}\right), \quad x>750$$ approximates the length of a home mortgage of \(\$ 150,000\) at \(6 \%\) in terms of the monthly payment. In the model, \(t\) is the length of the mortgage in years and \(x\) is the monthly payment in dollars. (a) Use the model to approximate the lengths of a \(\$ 150,000\) mortgage at \(6 \%\) when the monthly payment is \(\$ 897.72\) and when the monthly payment is \(\$ 1659.24\) (b) Approximate the total amounts paid over the term of the mortgage with a monthly payment of \(\$ 897.72\) and with a monthly payment of \(\$ 1659.24 .\) (c) Approximate the total interest charges for a monthly payment of \(\$ 897.72\) and for a monthly payment of \(\$ 1659.24\) (d) What is the vertical asymptote for the model? Interpret its meaning in the context of the problem.
Four-legged animals run with two different types of motion: trotting and galloping. An animal that is trotting has at least one foot on the ground at all times, whereas an animal that is galloping has all four feet off the ground at some point in its stride. The number of strides per minute at which an animal breaks from a trot to a gallop depends on the weight of the animal. Use the table to find a logarithmic equation that relates an animal's weight \(x\) (in pounds) and its lowest galloping speed \(y\) (in strides per minute). $$ \begin{array}{|c|c|} \hline \text { Weight, } x & \text { Galloping speed, } y \\ \hline 25 & 191.5 \\ 35 & 182.7 \\ 50 & 173.8 \\ 75 & 164.2 \\ 500 & 125.9 \\ 1000 & 114.2 \\ \hline \end{array} $$
Evaluate the logarithm using the change-of-base formula. Round your result to three decimal places. $$\log _{1 / 4} 5$$
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