Exponential and logarithmic functions - Examples

  • Examples
    Without using a calculator, find the values of the following expressions.
    (a) \(\frac{{\log 32 + \log 0.25}}{{\log 5 - 5\log 2 - 1}}\)

    (b) \(\frac{{2 + \log 16 - 2\log 0.04}}{{4\log \sqrt 5 - \log 0.25}}\)
  • Solutions
    \(\begin{array}{1}(a)\frac{{\log 32 + \log 0.25}}{{\log 5 - 5\log 2 - 1}}
    = \frac{{\log 32 + \log 0.25}}{{\log 5 - \log {2^5} - \log 10}}\\ = {\frac{{\log 32 + \log 0.25}}{{\log 5 - \log 32 - \log 10}}^{}}\\ = {\frac{{\log (32 \times 0.25)}}{{\log {\textstyle{5 \over {32{\rm{ }} \times {\rm{ }}10}}}}}^{}}\\ = {\frac{{\log 8}}{{\log {\textstyle{1 \over {64}}}}}^{}}\\ = {\frac{{\log 8}}{{\log {8^{ - 2}}}}^{}}\\ = {\frac{{\log 8}}{{ - {\rm{ }}2\log 8}}^{}}\\ = {\underline{\underline { - {\rm{ }}\frac{1}{2}}} ^{}}\end{array}\)

    \(\begin{array}{1}(b)\frac{{2 + \log 16 - 2\log 0.04}}{{4\log \sqrt 5 - \log 0.25}}
    = \frac{{\log 100 + \log 16 - \log {{({\textstyle{4 \over {100}}})}^2}}}{{\log {5^2} - \log {\textstyle{{25} \over {100}}}}}\\ = {\frac{{\log {\textstyle{{100{\rm{ }} \times {\rm{ }}16} \over {{\textstyle{{16} \over {10{\rm{ }}000}}}}}}}}{{\log {\textstyle{{25} \over {{\textstyle{{25} \over {100}}}}}}}}^{}}\\ = {\frac{{\log 1{\rm{ }}000{\rm{ }}000}}{{\log 100}}^{^{}}}\\ = \frac{6}{2}\\ = {\underline{\underline 3} ^{}}\end{array}\)

  • Examples
    If α and β are the two roots of the equation \(2{x^2} - 50x + 20 = 0\), find the value of \(\log \alpha + \log \beta \).
  • Solutions
    \(\begin{array}{1}2{x^2} - 50x + 20 = 0\\\alpha \beta = Product   Of   Roots\\
    = \frac{{20}}{2}\\ = 10\end{array}\)
    ∴ \(\begin{array}{1}\log \alpha + \log \beta = \log \alpha \beta \\ = \log 10\\ = \underline{\underline {{\rm{ }}1{\rm{ }}}} \end{array}\)

  • Examples
    Solve the equation \(\log (3{x^2} - 16) = \log (2x - 11)\).
    (Give your answer correct to 3 significant figures if necessary.)
  • Solutions
    \(\log (3{x^2} - 16) = \log (2x - 11)\)
    Assume that \(3{x^2} - 16 > 0\) and \(2x - 11 > 0\).
    ∴ \(\begin{array}{c}3{x^2} - 16 = 2x - 11\\3{x^2} - 2x - 5 = 0\\(x + 1)(3x - 5) = {0^{}}\end{array}\)
    \(x = - 1\) or \(x = \frac{5}{3}\)

    When \(x = - 1\),
    \(\begin{array}{c}3{x^2} - 16 = 3{( - 1)^2} - 16\\ = 3 - {16^{}}\\ = - {13^{}}\\ < {0^{}}\end{array}\) and \(\begin{array}{c}2x - 11 = 2( - 1) - {11^{}}\\ = - 2 - {11^{}}\\ = - {13^{}}\\ < {0^{}}\end{array}\)

    When \(x = \frac{5}{3}\),
    \(\begin{array}{c}3{x^2} - 16 = 3{(\frac{5}{3})^2} - 16\\ = \frac{{25}}{3} - {16^{^{^{}}}}\\ = - \frac{{23}}{3}\\ < 0\end{array}\) and \(\begin{array}{c}2x - 11 = 2(\frac{5}{3}) - 11\\ = \frac{{10}}{3} - {11^{^{^{}}}}\\ = - \frac{{23}}{3}\\ < 0\end{array}\)

    ∵ When \(x = - 1\) and \(x = \frac{5}{3}\), both \(\log (3{x^2} - 16)\) and \(\log (2x - 11)\) are undefined.
    ∴ The equation has no solutions.