# IBDP Maths analysis and approaches Topic: SL 2.9 : Logarithmic functions and their graphs HL Paper 2

### Question

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 = A0e-kt where t ≥ 0 and A0 , k are positive constants.

At the time of death, a plant is defined to have 100 units of carbon-14.

(a) Show that A0 = 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. [3]

Ans:

(a) When $t=0$, $A=100$, i.e., we have $$\begin{eqnarray} 100 = A_0\text{e}^{-k\left(0\right)} \nonumber \\ A_0 = 100. \end{eqnarray}$$ (b) Let half-life be $t=5730$, then $$\begin{eqnarray} \frac{1}{2} = \text{e}^{-k\left(5730\right)} \nonumber \\ -\ln 2 = -k\left(5730\right) \nonumber \\ k = \frac{\ln 2}{5730}. \end{eqnarray}$$ (c) When $25%$ of carbon-14 has decayed, we are left with $A=75$. Thus, from the graphing calculator, solving $75=100\text{e}^{-\frac{\ln 2}{5730}t}$, we have $t\approx 2380\text{ years}$.

### Question

(a)     Find the solution of the equation

$\ln {2^{4x – 1}} = \ln {8^{x + 5}} + {\log _2}{16^{1 – 2x}},$

expressing your answer in terms of $$\ln 2$$.

(b)     Using this value of x, find the value of a for which $${\log _a}x = 2$$, giving your answer to three decimal places.

### Markscheme

(a)     rewrite the equation as $$(4x – 1)\ln 2 = (x + 5)\ln 8 + (1 – 2x){\log _2}16$$     (M1)

$$(4x – 1)\ln 2 = (3x + 15)\ln 2 + 4 – 8x$$     (M1)(A1)

$$x = \frac{{4 + 16\ln 2}}{{8 + \ln 2}}$$     A1

(b)     $$x = {a^2}$$     (M1)

$$a = 1.318$$     A1

Note: Treat 1.32 as an AP.

Award A0 for ±.

[6 marks]

### Question

(a) Let a = log2x, b = log2y, c = log2z. Write $$log_2(\frac{x^3\sqrt{y}}{z^4})$$ in terms of a, b and c.
(b) Let A = log2x, B = log4y, C = log8z. Write $$log_2(\frac{x^3\sqrt{y}}{z^4})$$ in terms of A, B and C.

Ans
(a) $$log_2(\frac{x^3\sqrt{y}}{z^4})=3a+\frac{1}{2}b-4c$$.
(b) $$log_2(\frac{x^3\sqrt{y}}{z^4})$$=3A+B-12C\)

### Question

Solve the simultaneous equations
$$log_2(y-1)=1+log_2x$$
$$2log_3y=2+log_3x$$

x = 1, y = 3 or $$x=\frac{1}{4}, y=\frac{3}{2}$$