Exponentials and logarithms – Year 12 A level maths test answers to download as PDF

← Back to the test

1. Answer to Question 1 (4 marks)

   The laws used are \(k\log_a x=\log_a x^k\), \(\log_a x+\log_a y=\log_a xy\) and \(\log_a x-\log_a y=\log_a\dfrac xy\).

   Use the power law \(2\log_53=\log_59\), then the multiplication and division laws.

   \(\log_5\dfrac{9\times50}{18}\) (2) \(=\log_525\) (1) \(=2\) (1).

   Since \(5^2=25\), the expression equals 2.

   Check with a calculator: \(2\log_53+\log_550-\log_518\approx1.365+2.431-1.796=2.000\) ✓.

   Do not write \(2\log_53\) as \(\log_56\): the power law squares the 3, it does not double it. The final mark needs the exact value 2, not a decimal approximation.

2. Answer to Question 2 (4 marks)

   Notice that \(4^x=(2^x)^2\), so substituting \(y=2^x\) gives a quadratic \(y^2=7y-12\).

   Let \(y=2^x\): \(y^2-7y+12=0\) (1), \((y-3)(y-4)=0\) (1). \(2^x=4\) gives \(x=2\) (1); \(2^x=3\) gives \(x=\log_23=1.58\) (1).

   So \(x=2\) or \(x=\log_23\approx1.58\). Both are valid because \(2^x\) can be 3 or 4.

   Check \(x=2\): \(4^2=16\) and \(7\times4-12=16\) ✓.

   Check \(x=\log_23\): \(4^x=(2^x)^2=9\) and \(7\times3-12=9\) ✓. The exact form \(x=\dfrac{\ln3}{\ln2}\) is equally acceptable for the final mark.

3. Answer to Question 3 (4 marks)

   Both \(2x+1\) and \(x-1\) must be positive for the logarithms to exist, so any solution must satisfy \(x\gt 1\).

   Combine with the multiplication law: \(\log_3[(2x+1)(x-1)]=2\), so \((2x+1)(x-1)=3^2\).

   \((2x+1)(x-1)=9\) (1), \(2x^2-x-10=0\) (1), \((2x-5)(x+2)=0\) (1). \(x=2.5\) since \(x=-2\) is invalid (1).

   \(x=-2\) is rejected because \(\log_3(x-1)\) would be the logarithm of a negative number. So \(x=2.5\).

   Check: \(\log_3(2\times2.5+1)+\log_3(2.5-1)=\log_36+\log_31.5=\log_39=2\) ✓.

   The last mark is withheld if \(x=-2\) is not explicitly rejected.

4. Answer to Question 4 (4 marks)

   Take logarithms of \(y=ab^x\) to get a straight line with intercept \(\log a\) and gradient \(\log b\).

   \(\log y=\log a+x\log b\) (1). \(\log a=1.3\), \(a=19.9\) (1). \(\log b=0.2\), \(b=1.58\) (2).

   So \(a=10^{1.3}=19.9\) and \(b=10^{0.2}=1.58\), giving the model \(y=19.9\times1.58^x\).

   Check: when \(x=0\), \(y=a=19.9\) and \(\log_{10}19.9=1.30\), which is the intercept ✓.

   A common slip is to take \(a=1.3\) instead of \(10^{1.3}\): the intercept is \(\log_{10}a\), not \(a\) itself. Using \(\ln\) in place of \(\log_{10}\) here would give wrong values of both constants.

5. Answer to Question 5 (4 marks)

   1. Substitute \(t=4\). \(80e^{-0.6}=43.9\) mg (1) After 4 hours the mass is 43.9 mg. Check: \(e^{-0.6}\approx0.549\) and \(80\times0.549\approx43.9\) ✓.
   2. The half-life is the time for the mass to fall from 80 to 40 mg. \(e^{-0.15t}=\tfrac12\), \(t=\dfrac{\ln2}{0.15}=4.62\) h (2) The half-life is 4.62 hours. This half-life does not depend on the initial dose of 80 mg.
   3. \(\tfrac5{80}=\tfrac1{16}\); take logarithms and reverse the inequality. \(e^{-0.15t}\lt \tfrac1{16}\), \(t\gt \dfrac{\ln16}{0.15}=18.5\) h (1) The mass falls below 5 mg after about 18.5 hours. Note that 5 mg is \(\tfrac1{16}\) of 80 mg, so this is exactly four half-lives: \(4\times4.62=18.5\) h ✓.

Total : 20 marks