Exponential and logarithmic functions - Examples

  • Examples
    The magnitude of an earthquake is 6.8 in the Richter scale, and the energy released from an subsequent earthquake is 1% of that released from the first earthquake. Find the magnitude of the subsequent earthquake. (Give your answer correct to 1 decimal place.)
    [ The relation between the magnitude (M) of an earthquake and the energy released (E units) from the earthquake can be expressed as \(M = \frac{2}{3}\log E - 2.9\). ]
  • Solutions
    Let \({E_0}\) units and E units be the energy released from the first earthquake and that released from the subsequent earthquake respectively.
    \(\begin{array}{1}6.8 = \frac{2}{3}\log {E_0} - 2.9\\\frac{2}{3}\log {E_0} = 9.7\\\log {E_0} = 14.55\\{E_0} = {10^{14.55}}\end{array}\)
    ∵ \(E = {E_0} \times \frac{1}{{100}}\)
    \(\begin{array}{1}∴E = {10^{14.55}} \times \frac{1}{{100}}\\ = {10^{14.55}} \times {10^{ - 2}}\\ = {10^{12.55}}\end{array}\)
    Magnitude of the subsequent earthquake
    \(\begin{array}{l} = \frac{2}{3}\log {10^{12.55}} - 2.9\\ = \frac{2}{3} \times 12.55 - 2.9\end{array}\)
    \( = 5.5\) (corr. to 1 d.p.)
    ∴ The magnitude of the subsequent earthquake is 5.5 in the Richter scale.

  • Examples
    Solve the equation \({\log _4}({x^2} + 3) = {\log _4}x + 1\). (Give your answer correct to 3 significant figures if necessary.)
  • Solutions
    Assume that \({x^2} + 3 > 0\) and \(x > 0\).
    \(\begin{array}{1}{\log _4}({x^2} + 3) = {\log _4}x + 1\\{\log _4}({x^2} + 3) = {\log _4}x + {\log _4}4\\{\log _4}({x^2} + 3) = {\log _4}4x\end{array}\)
    \(\begin{array}{1}{x^2} + 3 = 4x\\{x^2} - 4x + 3 = 0\\(x - 3)(x - 1) = 0\end{array}\)
    \(x = 3\) or \(x = 1\)
    When \(x = 3\),
    \(\begin{array}{1}{x^2} + 3 = {3^2} + 3\\ = 9 + 3\\ = 12\\ > 0\end{array}\) and \(\begin{array}{c}4x = 4{(3)^{}}\\ = 12\\ > 0\\\end{array}\)
    When \(x = 1\),
    \(\begin{array}{1}{x^2} + 3 = {1^2} + 3\\ = 1 + 3\\ = 4\\ > 0\end{array}\) and \(\begin{array}{c}4x = 4{(1)^{}}\\ = 4\\ > 0\\\end{array}\)
    ∵ \(x = \underline{\underline 3} \) or \(x = \underline{\underline {{\rm{ }}1{\rm{ }}}} \)

  • Examples
    The figure below shows the graph of \(y = {\log _{\frac{1}{7}}}x\).
    [GraphMissing:Latex_exponential and logarithmic funtions 02 Q110]
    (a) Using the given graph, find the values of the following. (Give your answers correct to 1 decimal place if necessary.)
    (i) \({\log _{\frac{1}{7}}}0.8\)
    (ii) \({\log _{\frac{1}{7}}}1.5\)
    (iii) \({\log _{\frac{1}{7}}}3.2\)
    (b) Solve the following inequalities graphically. (Give your answers correct to 1 decimal place if necessary.)
    (i) \({\log _{\frac{1}{7}}}x < 0.6\)
    (ii) \({\log _{\frac{1}{7}}}x \ge - 0.3\)
  • Solutions
    [GraphMissing:Latex_exponential and logarithmic funtions 02 Q110]
    (a) From the graph,
    (i) \({\log _{\frac{1}{7}}}0.8 = \underline{\underline {0.1}} \) (corr. to 1 d.p.)
    (ii) \({\log _{\frac{1}{7}}}1.5 = \underline{\underline { - {\rm{ }}0.2}} \) (corr. to 1 d.p.)
    (iii) \({\log _{\frac{1}{7}}}3.2 = \underline{\underline { - {\rm{ }}0.6}} \) (corr. to 1 d.p.)

    (b) From the graph,
    (i) the solutions of the inequality \({\log _{\frac{1}{7}}}x < 0.6\) are \(x > 0.3\) (corr. to 1 d.p.).
    (ii)the solutions of the inequality \({\log _{\frac{1}{7}}}x \ge - 0.3\) are \(0 < x \le 1.8\) (corr. to 1 d.p.).