IB DP Math AA: Topic: SL 1.7 :Laws of exponents and Log HL Paper 2

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]

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