Chapter 5: Problem 27
Evaluate the given expression without using a calculator. $$\ln e^{15}$$
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Chapter 5: Problem 27
Evaluate the given expression without using a calculator. $$\ln e^{15}$$
These are the key concepts you need to understand to accurately answer the question.
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The U.S. Department of Commerce estimated that there were 54 million Internet users in the United States in 1999 and 85 million in 2002 . (a) Find an exponential function that models the number of Internet users in year \(x,\) with \(x=0\) corresponding to 1999 (b) For how long is this model likely to remain accurate? [Hint: The current U.S. population is about 305 million.]
Between 1790 and \(1860,\) the population y of the United States (in millions) in year x was given by \(y=3.9572\left(1.0299^{\circ}\right),\) where \(x=0\) corresponds to \(1790 .\)F ind the U.S. population in the given year. $$1800$$
Rationalize the denominator and simplify your answer. $$\frac{1+\sqrt{3}}{5+\sqrt{10}}$$
A bicycle store finds that the number \(N\) of bikes sold is related to the number \(d\) of dollars spent on advertising by \(N=51+100 \cdot \ln (d / 100+2)\) (a) How many bikes will be sold if nothing is spent on advertising? If \(\$ 1000\) is spent? If \(\$ 10,000\) is spent? (b) If the average profit is \(\$ 25\) per bike, is it worthwhile to spend \(\$ 1000\) on advertising? What about \(\$ 10,000 ?\) (c) What are the answers in part (b) if the average profit per bike is \(\$ 35 ?\)
The population of St. Petersburg, Florida (in thousands) can be approximated by the function $$ g(x)=-127.9+81.91 \ln x \quad(x \geq 70) $$ where \(x=70\) corresponds to 1970 (a) Estimate the population in 1995 and 2003 . (b) If this model remains accurate, when will the population be \(260,000 ?\)
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