6.4E: Exercises
- Page ID
- 120176
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Exercises
In Exercises 1 - 24, solve the equation analytically.
- \(\log(3x-1) = \log(4-x)\)
- \(\log_{2}\left(x^{3}\right) = \log_{2}(x)\)
- \(\ln\left(8-x^2\right)=\ln(2-x)\)
- \(\log_{5}\left(18-x^2\right) = \log_{5}(6-x)\)
- \(\log_{3}(7-2x) = 2\)
- \(\log_{\frac{1}{2}} (2x-1) = -3\)
- \(\ln\left(x^2-99\right) = 0\)
- \(\log(x^2-3x) = 1\)
- \(\log_{125} \left(\dfrac{3x-2}{2x+3}\right)=\dfrac{1}{3}\)
- \(\log\left(\dfrac{x}{10^{-3}}\right) = 4.7\)
- \(-\log(x) = 5.4\)
- \(10\log\left(\dfrac{x}{10^{-12}}\right) = 150\)
- \(6-3\log_{5}(2x)=0\)
- \(3\ln(x)-2=1-\ln(x)\)
- \(\log_{3}(x - 4) + \log_{3}(x + 4) = 2\)
- \(\log_{5}(2x + 1) + \log_{5}(x + 2) = 1\)
- \(\log_{169}(3x + 7) - \log_{169}(5x - 9) = \dfrac{1}{2}\)
- \(\ln(x+1) - \ln(x) = 3\)
- \(2\log_{7}(x) = \log_{7}(2) + \log_{7}(x+12)\)
- \(\log(x) - \log(2) = \log(x+8) - \log(x+2)\)
- \(\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8\)
- \(\ln(\ln(x)) = 3\)
- \(\left(\log(x)\right)^2=2\log(x)+15\)
- \(\ln(x^{2}) = (\ln(x))^{2}\)
In Exercises 25 - 30, solve the inequality analytically.
- \(\dfrac{1 - \ln(x)}{x^{2}} < 0\)
- \(x\ln(x) - x > 0\)
- \(10\log\left(\dfrac{x}{10^{-12}}\right) \geq 90\)
- \(5.6 \leq \log\left(\dfrac{x}{10^{-3}}\right) \leq 7.1\)
- \(2.3 < -\log(x) < 5.4\)
- \(\ln(x^{2}) \leq (\ln(x))^{2}\)
In Exercises 31 - 34, use your calculator to help you solve the equation or inequality.
- \(\ln(x) = e^{-x}\)
- \(\ln(x) = \sqrt[4]{x}\)
- \(\ln(x^{2} + 1) \geq 5\)
- \(\ln(-2x^{3} - x^{2} + 13x - 6) < 0\)
- Since \(f(x) = e^{x}\) is a strictly increasing function, if \(a < b\) then \(e^{a} < e^{b}\). Use this fact to solve the inequality \(\ln(2x + 1) < 3\) without a sign diagram. Use this technique to solve the inequalities in Exercises 27 - 29. (Compare this to Exercise 46 in Section 6.3.)
- Solve \(\ln(3 - y) - \ln(y) = 2x + \ln(5)\) for \(y\).
- In Example 6.4.4 we found the inverse of \(f(x) = \dfrac{\log(x)}{1-\log(x)}\) to be \(f^{-1}(x) = 10^{\frac{x}{x+1}}\).
- Show that \(\left(f^{-1} \circ f\right)(x) = x\) for all \(x\) in the domain of \(f\) and that \(\left(f \circ f^{-1}\right)(x) = x\) for all \(x\) in the domain of \(f^{-1}\).
- Find the range of \(f\) by finding the domain of \(f^{-1}\).
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Let \(g(x) = \dfrac{x}{1 - x}\) and \(h(x) = \log(x)\). Show that \(f = g \circ h\) and \((g \circ h)^{-1} = h^{-1} \circ g^{-1}\).
(We know this is true in general by Exercise 31 in Section 5.2, but it’s nice to see a specific example of the property.)
- Let \(f(x) = \dfrac{1}{2}\ln\left(\dfrac{1 + x}{1 - x}\right)\). Compute \(f^{-1}(x)\) and find its domain and range.
- Explain the equation in Exercise 10 and the inequality in Exercise 28 above in terms of the Richter scale for earthquake magnitude. (See Exercise 75 in Section 6.1.)
- Explain the equation in Exercise 12 and the inequality in Exercise 27 above in terms of sound intensity level as measured in decibels. (See Exercise 76 in Section 6.1.)
- Explain the equation in Exercise 11 and the inequality in Exercise 29 above in terms of the pH of a solution. (See Exercise 77 in Section 6.1.)
- With the help of your classmates, solve the inequality \(\sqrt[n]{x} > \ln(x)\) for a variety of natural numbers \(n\). What might you conjecture about the “speed” at which \(f(x) = \ln(x)\) grows versus any principal \(n^{\textrm{th}}\) root function?
Answers
- \(x = \frac{5}{4}\)
- \(x = 1\)
- \(x=-2\)
- \(x=-3,\, 4\)
- \(x=-1\)
- \(x=\frac{9}{2}\)
- \(x=\pm 10\)
- \(x=-2,\, 5\)
- \(x = -\frac{17}{7}\)
- \(x = 10^{1.7}\)
- \(x = 10^{-5.4}\)
- \(x = 10^{3}\)
- \(x=\frac{25}{2}\)
- \(x=e^{3/4}\)
- \(x = 5\)
- \(x = \frac{1}{2}\)
- \(x = 2\)
- \(x = \frac{1}{e^3-1}\)
- \(x=6\)
- \(x=4\)
- \(x = 81\)
- \(x = e^{e^3}\)
- \(x=10^{-3}, \, 10^{5}\)
- \(x = 1, \, x = e^{2}\)
- \((e, \infty)\)
- \((e, \infty)\)
- \(\left[10^{-3}, \infty \right)\)
- \(\left[10^{2.6}, 10^{4.1}\right]\)
- \(\left(10^{-5.4}, 10^{-2.3}\right)\)
- \((0, 1] \cup [e^{2}, \infty)\)
- \(x \approx 1.3098\)
- \(x \approx 4.177, \, x \approx 5503.665\)
- \(\approx (-\infty, -12.1414) \cup (12.1414, \infty)\)
- \(\approx (-3.0281, -3) \cup (0.5, 0.5991) \cup (1.9299, 2)\)
- \(-\dfrac{1}{2} < x < \dfrac{e^{3} - 1}{2}\)
- \(y = \dfrac{3}{5e^{2x} + 1}\)
- \(f^{-1}(x) = \dfrac{e^{2x} - 1}{e^{2x} + 1} = \dfrac{e^{x} - e^{-x}}{e^{x} + e^{-x}}\). The domain of \(f^{-1}\) is \((-\infty, \infty)\) and its range is the same as the domain of \(f\), namely \((-1, 1)\).