In Exercises solvelogeqexfirst - solvelogeqexlast, solve the equation analytically.

\(\log (3x-1) = \log (4-x)\)

\(x = \answer {\frac {5}{4}}\)

\(\log _{2}\left (x^{3}\right ) = \log _{2}(x)\)

\(x = \answer {1}\)

\(\ln \left (8-t^2\right )=\ln (2-t)\)

\(t=\answer {-2}\)

\(\log _{5}\left (18-t^2\right ) = \log _{5}(6-t)\)
\(t=-3\) \(t=-2\) \(t=1\) \(t=2\) \(t=4\) \(t=6\)
\(\log _{3}(7-2x) = 2\)

\(x=\answer {-1}\)

\(\log _{\frac {1}{2}} (2x-1) = -3\)

\(x=\answer {\frac {9}{2}}\)

\(\ln \left (t^2-99\right ) = 0\)
\(t=-33\) \(t=-10\) \(t=-9\) \(t=9\) \(t=10\) \(t=33\)
\(\log (t^2-3t) = 1\)
\(t=-3\) \(t=-2\) \(t=1\) \(t=5\) \(t=6\) \(t=8\)
\(\log _{125} \left (\frac {3x-2}{2x+3}\right )=\frac {1}{3}\)

\(x = \answer {-\frac {17}{7}}\)

\(\log \left (\frac {x}{10^{-3}}\right ) = 4.7\)

\(x = 10^{1.7}\)
\(-\log (x) = 5.4\)

\(x = 10^{-5.4}\)
\(10\log \left (\frac {x}{10^{-12}}\right ) = 150\)

\(x = 10^{3}\)
\(6-3\log _{5}(2t)=0\)

\(t=\answer {\frac {25}{2}}\)

\(3\ln (t)-2=1-\ln (t)\)

\(t=e^{3/4}\)
\(\log _{3}(t - 4) + \log _{3}(t + 4) = 2\)
\(t=-\sqrt {7}\) \(t=-5\) \(t=5\) \(t=\sqrt {7}\)
\(\log _{5}(2t + 1) + \log _{5}(t + 2) = 1\)

\(t = \answer {\frac {1}{2}}\)

\(\log _{169}(3x + 7) - \log _{169}(5x - 9) = \frac {1}{2}\)

\(x=\answer {2}\)

\(\ln (x+1) - \ln (x) = 3\)

\(x = \frac {1}{e^3-1}\)
\(2\log _{7}(t) = \log _{7}(2) + \log _{7}(t+12)\)

\(t=\answer {6}\)

\(\log (t) - \log (2) = \log (t+8) - \log (t+2)\)

\(t=\answer {4}\)

\(\log _{3}(x) = \log _{\frac {1}{3}}(x) + 8\)

\(x = \answer {81}\)

\(\ln (\ln (x)) = 3\)

\(x = e^{e^3}\)
\(\left (\log (t)\right )^2=2\log (t)+15\)
\(t=10^{-5}\) \(t=10^{-3}\) \(t=10^{3}\) \(t=10^{5}\)
\(\ln (t^{2}) = (\ln (t))^{2}\)

\(t = 1, \, x = e^{2}\)

In Exercises solvelogineqexfirst - solvelogineqexlast, solve the inequality analytically.

\(\frac {1 - \ln (t)}{t^{2}} < 0\)

\((e, \infty )\)
\(t\ln (t) - t > 0\)

\((e, \infty )\)
\(10\log \left (\frac {x}{10^{-12}}\right ) \geq 90\)

\((10^{-3}, \infty )\)
\(5.6 \leq \log \left (\frac {x}{10^{-3}}\right ) \leq 7.1\)

\(\left [10^{2.6}, 10^{4.1}\right ]\)
\(2.3 < -\log (x) < 5.4\)

\(\left (10^{-5.4}, 10^{-2.3}\right )\)
\(\ln (t^{2}) \leq (\ln (t))^{2}\)

\((0, 1] \cup [e^{2}, \infty )\)
In Exercises logeqcalcexfirst - logeqcalcexlast, use a graphing utility to help you solve the equation or inequality.
\(\ln (t) = e^{-t}\)

\(t\approx \answer {1.3098}\)

\(\ln (x) = \sqrt [4]{x}\)

\(x \approx 4.177, \, x \approx 5503.665\)
\(\ln (t^{2} + 1) \geq 5\)

\(\approx (-\infty , -12.1414) \cup (12.1414, \infty )\)
\(\ln (-2x^{3} - x^{2} + 13x - 6) < 0\)

\(\approx (-3.0281, -3) \cup (0.5, 0.5991) \cup (1.9299, 2)\)
In Exercises domaincomplicatedlogfirst - domaincomplicatedloglast, find the domain of the function.
\(r(x) = \frac {x}{1 - \ln (x)}\)

\((-\infty , e) \cup (e, \infty )\)
\(R(x) = \frac {x \ln (x)}{1 - \ln (x)}\)

\((0,e) \cup (e, \infty )\)
\(s(t) = \sqrt {2 - \log (t)}\)

\((0, 100]\)
\(c(t) = (2 \ln (t) -1)^{\frac {2}{3}}\)

\((0, \infty )\)
\(\ell (t) = \ln ( \ln (t))\)

\((1, \infty )\)
\(L(x) = \log \left ( \frac {x \ln (x)}{1 - \ln (x)} \right )\)

\((1,e)\)
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 sixfourdecibelineq - sixfourpHineq. (Compare this to Exercise onetoonelogexercise in Section ExponentialEquationsandInequalities.)

\(-\frac {1}{2} < x < \frac {e^{3} - 1}{2}\), so \(\left ( -\frac {1}{2}, \frac {e^{3} - 1}{2}\right )\)
Solve \(\ln (3 - y) - \ln (y) = 2x + \ln (5)\) for \(y\).

\(y = \frac {3}{5e^{2x} + 1}\)
In Example logfracinverse we found the inverse of \(f(x) = \frac {\log (x)}{1-\log (x)}\) to be \(f^{-1}(x) = 10^{\frac {x}{x+1}}\).
  1. Algebraically check our answer by verifying \(\left (f^{-1} \circ f\right )(x) = x\) 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 \(f^{-1}\).
  2. Find the range of \(f\) by finding the domain of \(f^{-1}\).

  3. Let \(g(x) = \frac {x}{1 - x}\) and \(h(x) = \log (x)\). Show that \(f = g \circ h\) and \((g \circ h)^{-1} = h^{-1} \circ g^{-1}\).

    NOTE: We know this is true in general by Exercise fcircginverse in Section InverseFunctions, but it’s nice to see a specific example of the property.

Let \(f(x) = \frac {1}{2}\ln \left (\frac {1 + x}{1 - x}\right )\). Compute \(f^{-1}(x)\) and find its domain and range.

\(f^{-1}(x) = \frac {e^{2x} - 1}{e^{2x} + 1} = \frac {e^{x} - e^{-x}}{e^{x} + e^{-x}}\).

To see why we rewrite this in this form, see Exercise andtheresthyperbolic in Section ParametricEquations.

The domain of \(f^{-1}\) is \((-\infty , \infty )\) and its range is the same as the domain of \(f\), namely \((-1, 1)\).

Explain the equation in Exercise sixfourRichterequ and the inequality in Exercise sixfourRichterineq above in terms of the Richter scale for earthquake magnitude. (See Exercise Richterexercise in Section ExponentialFunctions.)
Explain the equation in Exercise sixfourdecibelequ and the inequality in Exercise sixfourdecibelineq above in terms of sound intensity level as measured in decibels. (See Exercise decibelexercise in Section ExponentialFunctions.)
Explain the equation in Exercise sixfourpHequ and the inequality in Exercise sixfourpHineq above in terms of the pH of a solution. (See Exercise pHexercise in Section ExponentialFunctions.)
  1. With the help of your classmates, numerically and graphically investigate \(\ds {\lim _{x \rightarrow \infty }}\) \(\frac {\ln (x)}{x^{p}}\) for various real number powers, \(p > 0\).
  2. With the help of your classmates, numerically and graphically investigate \(\ds {\lim _{x \rightarrow 0^{+}}}\) \(x^{p} \, \ln (x) \) for various real number powers, \(p > 0\).
  3. What do numericalinvestigationlimitlnxoverx and numericalinvestigationlimitlnxtimesx suggest about the relative growth rates of powers of \(x\) and \(\ln (x)\)?
In Exercises logcurvesketchfirst - logcurvesketchlast a function \(f\) along with its derivatives \(f'\) and \(f''\) are given.
  • Find the domain of \(f\).
  • Find the \(x\)- and \(y\)-intercepts of the graph of each function, if any.
  • Use limits to determine the end behavior and behavior at the endpoints of the domain.
  • Use \(f'\) to determine the open intervals over which \(f\) is increasing or decreasing.
  • Determine the local extrema, if any.
  • Use \(f''\) to determine the open intervals over which the graph of \(f\) is concave up or concave down.
  • Determine the inflection points of the graph, if any.

\(f(x) = \ln (x) - \ln (5-x)\), \(f'(x) = \frac {1}{x} + \frac {1}{5-x}\), \(f''(x) = \frac {1}{(5-x)^2} - \frac {1}{x^2}\)

  • Domain: \((0, 5)\).
  • \(x\)-intercept: \(\left ( \frac {5}{2}, 0\right )\); there is no \(y\)-intercept.
  • \(\ds {\lim _{x \rightarrow 0^{+}} f(x) = -\infty }\), \(\ds {\lim _{x \rightarrow 5^{-}} f(x) = \infty }\); we have two vertical asymptotes: \(x = 0\) and \(x = 5\).
  • \(f\) is always increasing: \((0, 5)\).
  • There are no local extrema.
  • The graph of \(f\) is concave up on \(\left (0, \frac {5}{2}\right )\) and concave down on \(\left ( \frac {5}{2}, 5\right )\).
  • The inflection point is \(\left ( \frac {5}{2}, 0\right )\).
\(f(x) = \frac {\ln (x)}{x}\), \(f'(x) = \frac {1 - \ln (x)}{x^2}\), \(f''(x) = \frac {2 \ln (x) - 3}{x^3}\).

  • Domain: \((0, \infty )\).
  • \(x\)-intercept: \(\left ( 1 , 0\right )\); there is no \(y\)-intercept.
  • \(\ds {\lim _{x \rightarrow 0^{+}} f(x) = -\infty }\), \(\ds {\lim _{x \rightarrow \infty } f(x) = 0}\); we have a vertical and horizontal asymptote: \(x = 0\) and \(y = 0\).
  • \(f\) is increasing on \((0, e)\) and decreasing on \((e, \infty )\).
  • There is a local (absolute) max at \(\left (e, \frac {1}{e}\right )\).
  • The graph of \(f\) is concave up on \(\left (e^{\frac {3}{2}}, \infty \right )\) and concave down on \(\left (0 , e^{\frac {3}{2}} \right )\).
  • The inflection point is \(\left ( e^{\frac {3}{2}}, \frac {3}{2 e^{\frac {3}{2}}} \right )\).