In Exercises expeqnfirst - expeqnlast, solve the equation analytically.
\(2^{4x} = 8\)

\(x = \answer {\frac {3}{4}}\)

\(3^{(x - 1)} = 27\)

\(x = \answer {4}\)

\(5^{2x-1} = 125\)

\(x=\answer {2}\)

\(4^{2t} = \frac {1}{2}\)

\(t = \answer {-\frac {1}{4}}\)

\(8^{t} = \frac {1}{128}\)

\(t = \answer {-\frac {7}{3}}\)

\(2^{(t^{3} - t)} = 1\)
\(t=-2\) \(t=-1\) \(t=0\) \(t=1\) \(t=2\) \(t=3\) \(t=8\)
\(3^{7x} = 81^{4-2x}\)

\(x = \answer {\frac {16}{15}}\)

\(9 \cdot 3^{7x} = \left (\frac {1}{9}\right )^{2x}\)

\(x=\answer {-\frac {2}{11}}\)

\(3^{2x} = 5\)

\(x = \dfrac {\ln (5)}{2\ln (3)}\)
\(5^{-t} = 2\)

\(t = -\dfrac {\ln (2)}{\ln (5)}\)
\(5^{t} = -2\)

No solution.
\(3^{(t - 1)} = 29\)

\(t = \dfrac {\ln (29) + \ln (3)}{\ln (3)}\)
\((1.005)^{12x} = 3\)

\(x = \dfrac {\ln (3)}{12\ln (1.005)}\)
\(e^{-5730k} = \frac {1}{2}\)

\(k = \dfrac {\ln \left (\frac {1}{2}\right )}{-5730} = \dfrac {\ln (2)}{5730} \)
\(2000e^{0.1t} = 4000\)

\(t=\dfrac {\ln (2)}{0.1} = 10\ln (2)\)
\(500\left (1-e^{2t}\right ) = 250\)

\(t=\frac {1}{2}\ln \left (\frac {1}{2}\right ) = -\frac {1}{2}\ln (2)\)
\(70 + 90e^{-0.1t} = 75\)

\(t = \dfrac {\ln \left (\frac {1}{18}\right )}{-0.1} =10 \ln (18)\)
\(30-6e^{-0.1t}=20\)

\(t=-10\ln \left (\frac {5}{3}\right ) = 10\ln \left (\frac {3}{5}\right )\)
\(\dfrac {100e^{x}}{e^{x}+2}=50\)

\(x=\answer {\ln (2)}\)

\(\dfrac {5000}{1+2e^{-3t}}=2500\)

\(t=\frac {1}{3}\ln (2)\)
\(\dfrac {150}{1 + 29e^{-0.8t}} = 75\)

\(t = \dfrac {\ln \left (\frac {1}{29}\right )}{-0.8} = \dfrac {5}{4}\ln (29)\)
\(25\left (\frac {4}{5}\right )^{x} = 10\)

\(x = \dfrac {\ln \left (\frac {2}{5}\right )}{\ln \left (\frac {4}{5}\right )} = \dfrac {\ln (2)-\ln (5)}{\ln (4) - \ln (5)}\)
\(e^{2x} = 2e^{x}\)

\(x = \answer {\ln (2)}\)

\(7e^{2t} = 28e^{-6t}\)

\(t = -\frac {1}{8} \ln \left (\frac {1}{4} \right ) = \frac {1}{4}\ln (2)\)
\(3^{(x - 1)} = 2^{x}\)

\(x = \dfrac {\ln (3)}{\ln (3) - \ln (2)}\)
\(3^{(x - 1)} = \left (\frac {1}{2}\right )^{(x + 5)}\)

\(x = \dfrac {\ln (3) + 5\ln \left (\frac {1}{2}\right )}{\ln (3) - \ln \left (\frac {1}{2}\right )} = \dfrac {\ln (3)-5\ln (2)}{\ln (3)+\ln (2)}\)
\(7^{3+7x} = 3^{4-2x}\)

\(x = \dfrac {4 \ln (3) - 3 \ln (7)}{7 \ln (7) + 2 \ln (3)}\)
\(e^{2t} - 3e^{t}-10=0\)

\(t=\answer {\ln (5)}\)

\(e^{2t} = e^{t}+6\)

\(t=\answer {\ln (3)}\)

\(4^{t} + 2^{t} = 12\)

\(x=\dfrac {\ln (3)}{\ln (2)}\)
\(e^{x}-3e^{-x}=2\)

\(x=\answer {\ln (3)}\)

\(e^{x}+15e^{-x}=8\)

\(x=\ln (3)\), \(\ln (5)\)
\(3^{x}+25\cdot 3^{-x}=10\)

\(x=\dfrac {\ln (5)}{\ln (3)}\)
In Exercises expineqfirst - expineqlast, solve the inequality analytically.
\(e^{x} > 53\)

\((\ln (53), \infty )\)
\(1000\left (1.005\right )^{12t} \geq 3000\)

\(\left [\dfrac {\ln (3)}{12\ln (1.005)}, \infty \right )\)
\(2^{(x^{3} - x)} < 1\)

\((-\infty , -1) \cup (0, 1)\)
\(25\left (\dfrac {4}{5}\right )^{x} \geq 10\)

\(\left (-\infty , \dfrac {\ln \left (\frac {2}{5}\right )}{\ln \left (\frac {4}{5}\right )} \right ] = \left (-\infty , \dfrac {\ln (2)-\ln (5)}{\ln (4)-\ln (5)} \right ]\)
\(\dfrac {150}{1 + 29e^{-0.8t}} \leq 130\)

\(\left (-\infty , \dfrac {\ln \left (\frac {2}{377}\right )}{-0.8} \right ] = \left (-\infty , \frac {5}{4}\ln \left (\dfrac {377}{2}\right ) \right ]\)
\(70 + 90e^{-0.1t} \leq 75\)

\(\left [\dfrac {\ln \left (\frac {1}{18}\right )}{-0.1}, \infty \right ) = [10\ln (18), \infty )\)
\(e^{-x} - xe^{-x} \geq 0\)

\((-\infty , 1]\)
\((1-e^{t}) t^{-1} \leq 0\)

\((-\infty , 0) \cup (0, \infty )\)
In Exercises calcexpineqfirst - calcexpineqlast, use a graphing utility to help you solve the equation or inequality.
\(2^{x} = x^2\)

\(x \approx -0.76666, \, x = 2, \, x = 4\)
\(e^{t} = \ln (t) + 5\)

\(x \approx 0.01866, \, x \approx 1.7115\)
\(e^{\sqrt {x}} = x + 1\)

\(x = \answer {0}\)

\(e^{-2t}-te^{-t} \leq 0\)

\(\approx [0.567, \infty )\)
\(3^{(x - 1)} < 2^{x}\)

\(\approx (-\infty , 2.7095)\)
\(e^{t} < t^{3} - t\)

\(\approx (2.3217, 4.3717)\)
In Exercises domaincomplicatedexpfirst - domaincomplicatedexplast, find the domain of the function.
\(T(x) = \dfrac {e^{x} - e^{-x}}{e^{x} + e^{-x}}\)

\((-\infty , \infty )\)
\(C(x) = \dfrac {e^{x} + e^{-x}}{e^{x} - e^{-x}}\)

\((-\infty , 0) \cup (0, \infty )\)
\(s(t) = \sqrt {e^{2t} - 3}\)

\(\left ( \frac {1}{2} \ln (3), \infty \right )\)
\(c(t) = \sqrt [3]{e^{2t} - 3}\)

\((-\infty , \infty )\)
\(L(x) = \log \left ( 3 - e^{x} \right )\)

\((-\infty , \ln (3))\)
\(\ell (x) = \ln \left ( \dfrac {e^{2x}}{e^{x}-2} \right )\)

\((\ln (2), \infty ) \)
Since \(f(x) = \ln (x)\) is a strictly increasing function, if \(0 < a < b\) then \(\ln (a) < \ln (b)\). Use this fact to solve the inequality \(e^{(3x - 1)} > 6\) without a sign diagram. Use this technique to solve the inequalities in Exercises expineqfirst - expineqlast. (NOTE: Isolate the exponential function first!)

\(x > \frac {1}{3}(\ln (6) + 1)\), so \(\left (\frac {1}{3}(\ln (6) + 1), \infty \right )\)
Compute the inverse of \(f(x) = \dfrac {e^{x} - e^{-x}}{2}\). State the domain and range of both \(f\) and \(f^{-1}\).

\(f^{-1} = \ln \left (x + \sqrt {x^{2} + 1}\right )\). Both \(f\) and \(f^{-1}\) have domain \((-\infty , \infty )\) and range \((-\infty , \infty )\).
In Example expfracinverse, we found that the inverse of \(f(x) = \dfrac {5e^{x}}{e^{x}+1}\) was \(f^{-1}(x) = \ln \left (\dfrac {x}{5-x}\right )\) but we left a few loose ends for you to tie up.
  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. With help of a graphing utility, graph \(y = f(x)\), \(y = f^{-1}(x)\) and \(y = x\) on the same set of axes. How does this help to verify our answer?

  4. Let \(g(x) = \dfrac {5x}{x+1}\) and \(h(x) = e^{x}\). Show that \(f = g \circ h\) and that \((g \circ h)^{-1} = h^{-1} \circ g^{-1}\).

    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.

  1. With the help of your classmates, numerically and graphically investigate \(\ds {\lim _{x \rightarrow \infty }}\) \(\frac {x^{p}}{e^{x}}\) for various real number powers, \(p\).
  2. What does part numericalinvestigationlimitxpoverex suggest about the relative growth rates of powers of \(x\) as opposed to \(e^{x}\)?
  3. For each power \(p\) you investigated in part numericalinvestigationlimitxpoverex, solve the inequality: \(\frac {x^{p}}{e^{x}} < \frac {1}{x}\).

  4. Use your results from part numericalxpoverexinequ to show that for each real number \(p\) you investigated in part numericalinvestigationlimitxpoverex, there is a real number \(M\) so that if \(x > M\), \(0 < \frac {x^{p}}{e^{x}} < \frac {1}{x}\).

    Since \(\ds {\lim _{x \rightarrow \infty }}\) \(\frac {1}{x} = 0\), what do you conclude about \(\ds {\lim _{x \rightarrow \infty }}\) \(\frac {x^{p}}{e^{x}}\)?

    (This Exercise foreshadows the celebrated Squeeze Theorem, Theorem squeezeth which we’ll formally introduce in Section Sequences.)

In Exercises exponentialcurvesketchfirst - exponentialcurvesketchlast a function \(f\) along with its derivatives \(f'\) and \(f''\) are given.
  • Find the \(x\)- and \(y\)-intercepts of the graph of each function, if any.
  • Use limits to determine the end behavior.
  • 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.
\(f(x) = \dfrac {5}{1 + e^{-x}}\), \(f'(x) = \dfrac {5 e^{-x}}{\left (1 + e^{-x} \right )^2}\), \(f''(x) = \dfrac {5e^{-x}\left (e^{-x}-1\right )}{\left (1+e^{-x}\right )^3}\)
\(f(x) = e^{-x} - e^{-2x}\), \(f'(x) = 2e^{-2x} - e^{-x}\), \(f''(x) = e^{-x} - 4 e^{-2x}\)