Question 5. [Maximum mark: 7]
All living plants contain an isotope of carbon called carbon-14. When a plant dies, the isotope
decays so that the amount of carbon-14 present in the remains of the plant decreases. The
time since the death of a plant can be determined by measuring the amount of carbon-14 still
present in the remains.
The amount, A, of carbon-14 present in a plant t years after its death can be modelled by \(A=A_{0}e^{-kt}\) where \(t\geq 0\) and \(A_{0}\) ,k are positive constants.
At the time of death, a plant is defined to have 100 units of carbon-14.
(a) Show that \(A_{0} = 100\). [1]
The time taken for half the original amount of carbon-14 to decay is known to be 5730 years.
(b) Show that \(k=\frac{\ln 2}{5730} \) [3]
(c) Find, correct to the nearest 10 years, the time taken after the plant’s death for 25% of the carbon-14 to decay.
▶️Answer/Explanation
(a) \(100=A_{0}e^{0}\)
\(100=A_{0}\)
(b) correct substitution of values into exponential equation
\(500=100e^{-5730k}\) OR \(e^{-5730k}=\frac{1}{2}\)
EITHER
\(-5730K=\ln \frac{1}{2}\)
\(\ln \frac{1}{2}=-ln2\) OR \( -\ln \frac{1}{2}=\ln 2\)
OR
\(e^{5730k}=2\)
\(5730k=\ln 2\)
THEN
\(k=\frac{\ln 2}{5730}\)
(c) if 25% of the carbon-14 has decayed,75 remains i.e, 75 units remain \(75=100e^{-\frac{\ln 2}{5730}t}\)
EITHER
using an appropriate graph to attempt to solve for t
OR
manipulating logs to attempt to solve for t
\(\ln 0.75=-\frac{\ln 2}{5730}\)
y=2738.164…
THEN
t=2380 (years)(correct to the nearest 10 years)
Question
Solve the equation \(2 – {\log _3}(x + 7) = {\log _{\tfrac{1}{3}}}2x\) .
▶️Answer/Explanation
Markscheme
\({\log _3}\left( {\frac{9}{{x + 7}}} \right) = {\log _3}\frac{1}{{2x}}\) M1M1A1
Note: Award M1 for changing to single base, M1 for incorporating the 2 into a log and A1 for a correct equation with maximum one log expression each side.
\(x + 7 = 18x\) M1
\(x = \frac{7}{{17}}\) A1
[5 marks]
Question
Consider \(a = {\log _2}3 \times {\log _3}4 \times {\log _4}5 \times \ldots \times {\log _{31}}32\). Given that \(a \in \mathbb{Z}\), find the value of a.
▶️Answer/Explanation
Markscheme
\(\frac{{\log 3}}{{\log 2}} \times \frac{{\log 4}}{{\log 3}} \times \ldots \times \frac{{\log 32}}{{\log 31}}\) M1A1
\( = \frac{{\log 32}}{{\log 2}}\) A1
\( = \frac{{5\log 2}}{{\log 2}}\) (M1)
\( = 5\) A1
hence \(a = 5\)
Note: Accept the above if done in a specific base eg \({\log _2}x\).
[5 marks]