Partial Solutions


  1. There is no solution provided for this question.

    1. There are no restrictions on \(x\).

      Converting \(\log_8 \left(\frac{1}{16\sqrt{2}}\right)= x\) to exponential form we have

      \[\begin{align*} 8^x &= \frac{1}{16\sqrt{2}} \\ \left(2^3\right)^x &= 2^{-4}\left(2^{-\frac{1}{2}}\right) \end{align*}\]

      So, \( 3x=-\frac{9}{2}\), therefore \(x=-\frac{3}{2}\).

    2. Restrictions: \( 4x+6 \gt 0 \), so \( x \gt -\frac{3}{2} \).

      Converting \(\log_3 (4x+6)=2\) to exponential form we have

      \[\begin{align*} 3^2 &= 4x+6 \\ 4x &= 9-6 \end{align*}\]

      Therefore, \(x=\frac{3}{4} \).

    3. Restrictions: \( 3x-2 \gt 0 \) so \( x \gt \frac{2}{3} \).

      Converting \( \log (3x-2)=1 \) to exponential form, we have

      \[\begin{align*} 10^1 &= 3x-2 \\ 3x &= 12 \end{align*}\]

      Therefore, \( x=4 \).

    4. Restrictions: \( x \gt 0 \)

      Converting \( \log_x (16) = -2 \) to exponential form, we have

      \[\begin{align*} x^{-2} &= 16 \\ \left(x^{-2}\right)^{-\frac{1}{2}} &= \pm 16^{-\frac{1}{2}} \\ x &= \pm \frac{1}{\sqrt{16}} \end{align*}\]

      However \(x \gt 0\), therefore \( x=\frac{1}{4}\).

    5. Restrictions: \(x \gt 0\)

      Converting \(\log_x (64) = \frac{3}{2}\) to exponential form, we have

      \[\begin{align*} x^{\frac{3}{2}} &= 64 \\ \left(x^{\frac{3}{2}}\right)^{\frac{2}{3}} &= 64^{\frac{2}{3}} \\ x &= \sqrt[3]{64}^2 \end{align*}\]

      Therefore, \( x=16 \).

    6. Restrictions: \(x \gt 0\) and \(6-x \gt 0\), so \(0\lt x \lt 6\).

      Converting \(\log_x (6-x)=2\) to exponential form, we have

      \[\begin{align*} x^2 &= 6-x \\ x^2+x-6 &= 0 \\ (x+3)(x-2) &= 0 \\ x &= -3, 2 \end{align*}\]

      However \( 0 \lt x \lt 6 \), therefore \( x=2 \).

  3. There is no solution provided for this question.
    1. \[\begin{align*} \log_6 (18) + \log_6 (2) &= \log_6 (18\times2) \\ &= \log_6 (32) \\ &= 2 \end{align*}\]
    2. \[\begin{align*} \log_2 (56) - \log_2 (7) &= \log_2 \left(\frac{56}{7}\right) \\ &= \log_2(8) \\ &= 3 \end{align*}\]
    3. \[\begin{align*} \log (4) -4\log (2) - \log (25) &= \log (4) -\log (2^4) - \log (25) \\ &= \log\left(\frac{4}{16\times 25}\right) \\ &= \log\left(\frac{1}{100}\right) \\ &= -2 \end{align*}\]
    4. \[\begin{align*} 5^{2\log_5 (6)} &= 5^{\log_5 \left(6^2\right)} \\ &= 36 \end{align*}\]
    5. \[\begin{align*} 3\log_5 (10) - \log_5 (40) &= \log_5 \left(10^3\right) - \log_5 (40) \\ &= \log_5 \left(\frac{1000}{40}\right) \\ &= \log_5 (25) \\ &= 2 \end{align*}\]
    6. \[\begin{align*} \log_4 (24)- 2\log_4 (3) + \frac{1}{2}\log_4 (144) &= \log_4 (24) - \log_4 \left(3^2\right) + \log_4 \left(144^{\frac{1}{2}}\right) \\ &= \log_4 \left(\frac{24\times12}{9}\right) \\ &= \log_4 (32) \\ &= \frac{5}{2} \end{align*}\]
    7. \[\begin{align*} \frac{\log_3 (216)}{\log_3 (6)} &= \log_6(216) \\ &= 3 \end{align*}\]
    8. \[\begin{align*} 9^{\log_3(12)-\log_3(3)} &= \left(3^2\right)^{\log_3\left(\frac{12}{3}\right)} \\ &= 3^{2\log_3\left(4\right)} \\ &= 3^{\log_3\left(4^2\right)} \\ &= 3^{\log_3\left(16\right)} \\ &= 16 \end{align*}\]
    9. \[\begin{align*} 3\log (200) - \frac{1}{2}\log (64) &= \log \left(200^3\right) - \log \left(64^\frac{1}{2}\right) \\ &= \log \left(8000000\right) - \log \left(8\right) \\ &= \log \left(\frac{8000000}{8}\right) \\ &= \log (1000000) \\ &= 6 \end{align*}\]
  5. There is no solution provided for this question.
  6. Using a scientific calculator with a common (base 10) log function, we have:
    1. \( \log (185) \approx 2.26717\). Therefore, \(\log (185)\) is approximately \(2.267\) to three decimal places of accuracy.
    2. \(\log_3 (65)=\frac{\log(65)}{\log(3)}\approx 3.79969\). Therefore, \(\log_3 (65)\) is approximately \(3.800\) to three decimal places of accuracy.
    3. \(\log_8 (5)=\frac{\log(5)}{\log(8)}\approx 0.77398\). Therefore, \(\log_8 (5)\) is approximately \(0.774\) to three decimal places of accuracy.
    4. \(2\log_{\frac{1}{4}} (24)= \frac{2\log(24)}{\log\left(\frac{1}{4}\right)} \approx -4.58496\). Therefore, \(2\log_{\frac{1}{4}} (24)\) is approximately \(-4.585\) to three decimal places of accuracy.
  7. There is no solution provided for this question.

    1. Using the change of base formula,

      \[\begin{align*} \frac{1}{\log_a (c)} &= \dfrac{1}{\frac{\log_n (c)}{\log_n (a)}} \\ &= \frac{\log_n (a)}{\log_n (c)} \\ &= \log_c (a) \end{align*}\]

      as required.

      Alternate Approach:

      Let \(\log_c (a) = m\). Expressing this in exponential form we have \(c^m=a\) so \(c=a^{\frac{1}{m}}\). Now,

      \[\begin{align*} \frac{1}{\log_a (c)} &= \frac{1}{\log_a (a^\frac{1}{m})} \\ &= \frac{1}{1/m} \\ &= m \\ &= \log_c (a) \end{align*}\]

      as required.

    2. Let \(\log_a (b) = m\) and \(\log_b (c) = n\).

      Expressing these in exponential form we have \(a^m=b\) and \(b^n=c\).

      So \(c=b^n=\left(a^m\right)^n=a^{mn}\)

      Now,

      \[\begin{align*} \log_a (c) &= \log_a (a^{mn}) \\ &= mn \\ &= \log_a (b) \log_b(c) \end{align*}\]

      as required.

      Alternate Approach:

      Using the change of base formula,

      \[\begin{align*} \log_a (b)\log_b(c) &= \left(\frac{\log_n (b)}{\log_n (a)}\right) \left(\frac{\log_n (c)}{\log_n (b)}\right) \\ &= \frac{\log_n (c)}{\log_n (a)} \\ &= \log_a (c) \end{align*}\]

      as required.

  9. There is no solution provided for this question.
  10. Find \(g(x)\).\[\begin{align*} g(x) &= \log_2\left(\dfrac{f(x)}{1-f(x)}\right)\\ &= \log_2 f(x)-\log_2 \left(1-f(x)\right)\\ &= \log_2 \left(\dfrac{2^{\frac{1}{x}+1}}{1+2^{\frac{1}{x}+1}}\right)-\log_2 \left(1-\dfrac{2^{\frac{1}{x}+1}}{1+2^{\frac{1}{x}+1}}\right)\\ &= \log_2 \left(2^{\frac{1}{x}+1}\right)-\log_2 \left(1+2^{\frac{1}{x}+1}\right)-\log_2 \left(\dfrac{1+2^{\frac{1}{x}+1}-2^{\frac{1}{x}+1}}{1+2^{\frac{1}{x}+1}}\right)\\ &= \frac{1}{x}+1-\log_2 \left(1+2^{\frac{1}{x}+1}\right)-\log_2 \left(\dfrac{1}{1+2^{\frac{1}{x}+1}}\right)\\ &= \frac{1}{x}+1-\log_2 \left(1+2^{\frac{1}{x}+1}\right)-\left(\log_2 (1)-\log_2 \left(1+2^{\frac{1}{x}+1}\right)\right)\\ &= \frac{1}{x}+1-\log_2 (1)\\ &= \frac{1}{x}+1 \end{align*}\] Now,\[\begin{align*} xg(x)-g\left(\frac{1}{x}\right) &= x\left(\frac{1}{x}+1\right)-\left(\dfrac{1}{\frac{1}{x}}+1\right)\\ &= 1+x-\left(x+1\right)\\ &= 0 \end{align*}\] as required.