Show that $A_0=100$.

Show that $k=\frac{\ln 2}{5730}$.

Find, correct to the nearest $10$ years, the time taken after the plant’s death for $25\%$ of the carbon-14 to decay.

$100=A_0{\text{e}}^0$             A1

$A_0=100$             AG

[1 mark]

correct substitution of values into exponential equation             (M1)

$50=100{\text{e}}^{-5730k}$  OR  ${\text{e}}^{-5730k}=\frac{1}{2}$

EITHER

$-5730k=\ln \frac{1}{2}$             A1

$\ln \frac{1}{2}=-\ln 2$  OR  $-\ln \frac{1}{2}=\ln 2$             A1

OR

${\text{e}}^{5730k}=2$             A1

$5730k=\ln 2$             A1

THEN

$k=\frac{\ln 2}{5730}$             AG

Note: There are many different ways of showing that $k=\frac{\ln 2}{5730}$ which involve showing different steps. Award full marks for at least two correct algebraic steps seen.

[3 marks]

if $25\%$ of the carbon-14 has decayed, $75\%$ remains ie, $75$ units remain                      (A1)

$75=100{\text{e}}^{-\frac{\ln 2}{5730}t}$

EITHER

using an appropriate graph to attempt to solve for $t$                      (M1)

OR

manipulating logs to attempt to solve for $t$                               (M1)

$\ln 0.75=-\frac{\ln 2}{5730}t$

$t=2378.164\dots$

THEN

$t=2380$ (years) (correct to the nearest 10 years)                   A1

[3 marks]