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15 (CIE 2015, s, paper 21, question 10)
The relationship between experimental values of two variables, $x$ and $y$, is given by $y=A b^{x}$, where $A$ and $b$ are constants.
(i) By transforming the relationship $y=A b^{x}$, show that plotting $\ln y$ against $x$ should produce a straight line graph. $\quad[2]$
(ii) The diagram below shows the results of plotting $\ln y$ against $x$ for 7 different pairs of values of variables, $x$ and $y .$ A line of best fit has been drawn.
By taking readings from the diagram, find the value of $A$ and of $b$, giving cach value correct to significant figure. $[4]$
(iii) Estimate the value of $y$ when $x=2.5$. [2]
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$\begin{aligned}&\begin{aligned}\text{(i) }y=& A b^{x} \\\ln y=& \ln A+x \ln b \\\text { Choose } &(1.5,10),(4,7.6) \\\text { gradient } &=\frac{7.6-10}{4-1.5}=-0.96 \\\ln y-10 &=-0.96(x-1.5) \\\ln y &=-0.96 x+11.44\qquad \cdots (1) \\y &=e^{-0.96 x} \cdot e^{11.44} \\&=\left(-e^{0.96}\right)^{x} \cdot 92967 \\&=(0.38)^{x}(90000) \\\therefore A &=90,000, \quad b=0.4 . \\\text { (ii) When } x &=2.5, \text { by }(1) \\\ln y &=-0.96(2.5)+11.44=9 \\y &=e^{9}=8103\end{aligned}\end{aligned}$
16 (CIE 2015, w, paper 11, question 7)
Two variables, $x$ and $y$, are such that $y=A x^{h}$, where $A$ and $b$ are constants. When $\ln y$ is plotted against $\ln x$, a straight line graph is obtained which passes through the points $(1.4,5.8)$ and $(2.2,6.0)$.
(i) Find the value of $A$ and of $b$. $[4]$
(ii) Calculate the value of $y$ when $x=5$. $[2]$
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$\begin{aligned}&y=A x^{b} \\&\ln y=\ln A+b \ln x \quad \ldots(1) \\&\text { Two points }(1.4,5.8),(2.2,6.0) \\&\text { gradient }=\frac{6.0-5.8}{2.2-1.4}=\frac{1}{4} \\&\text { line equation } \\&\qquad \begin{aligned}\ln y-6 &=\frac{1}{4}(\ln x-2.2) \\\ln y &=\frac{1}{4} \ln x+5.45\end{aligned} \\&\text { compare with equation (1), } \\&\qquad \begin{array}{l}b=\frac{1}{4}, \ln A=5.45 \Rightarrow A=e^{5.45} \\\text { When } x=5,-y=e^{5.45} \cdot 5^{\frac{1}{4}}=348\end{array}\end{aligned}$
17 (CIE 2015, w, paper 23, question 11)
The trees in a certain forest are dying because of an unknown virus.
The number of trees, $N$, surviving $t$ years after the onset of the virus is shown in the table below.
$\begin{array}{|c|c|c|cc|c|c|}\hline \text{t }& 1 & 2 & 3 & 4 & 5 & 6 \\\hline \text{N } & 2000 & 1300 & 890 & 590 & 395 & 260 \\\hline\end{array}$
The relationship between $N$ and $t$ is thought to be of the form $N=A b^{-t}$.
(i) Transform this relationship into straight line form.
(ii) Using the given data, draw this straight line on the grid below. $[3]$
(iii) Use your graph to estimate the value of $A$ and of $b$.
If the trees continue to die in the same way, find
(iv) the number of trecs surviving after 10 years,
(v) the number of years taken until there are only 10 trees surviving. [2]
$\begin{aligned}\text{(i) }&\log N=\log A-t \log b\\\text{(ii) }&\begin{array}[t]{|c|c|c|c|c|c|c|}\hline t & 1 & 2 & 3 & 4 & 5 & 6 \\\hline \log N & 3.30 & 3.11 & 2.95 & 2.77 & 2.60 & 2.41 \\\hline \end{array}\end{aligned}$
Choose two points $(1,3.30)$ and $(6,2.415)$.
$\begin{aligned}\text{(iii) }&\text { gradient }=-\log b=\frac{2.415-3.3}{5} \rightarrow b=1.5\\&\text { intercept }=\log A=3.47 \rightarrow A=2950\\\text{(iv) }&t=10 \rightarrow N=\frac{2950}{1.5^{10}}=51\\\text{(v) }&N=10 \rightarrow 1.5^{t}=295 \rightarrow t=\frac{\log 295}{\log 1.5}\\&=14 \text { years }\end{aligned}$
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