Sec 4 A Math: Linear Law, practice questions & worked solutions

(a)

(i)

Plot $\lg y$ against $x$ and draw a straight line graph to illustrate the information.

(ii) $$\begin{aligned} y &= Ak^x\\ \lg y &= \lg Ak^x\\ \lg y &= x\lg k+\lg A \end{aligned}$$ From the graph, gradient $=\dfrac{0.92-0.59}{4-2}=0.165$. $$\begin{aligned} \lg y\text{-intercept} &= 0.26,\\ \lg A &= 0.26\\ A &= 1.82\\ \lg k &= 0.165\\ k &= 1.46 \end{aligned}$$

(b)

$$\begin{aligned} y &= x^2(ax+b)\\ \frac{y}{x^2} &= ax+b \text{or alternative } \frac{y}{x^3}=a+\frac{b}{x} \end{aligned}$$ Plot $\dfrac{y}{x^2}$ (or $\dfrac{y}{x^3}$) on the vertical axis against $x$ (or $\dfrac{1}{x}$) on the horizontal axis.

Gradient of line $=a$ (or $b$).

Vertical intercept $\left(\dfrac{y}{x^2}\text{ intercept}\right)$ of line $=b$ (or $a$).

Solution 1

(a)

$$\begin{aligned} n &= n_0(2)^{kt}\\ \lg n &= \lg n_0 + (k\lg 2)t \end{aligned}$$ Draw a straight line of $\lg n$ against $t$.

Best fit line

(b)

$$\begin{aligned} \lg n_0 &= 3.48\\ n_0 &= 10^{3.48}\\ &= 3020 \end{aligned}$$

(c)

$$\begin{aligned} k\lg 2 &= \frac{3.584 - 3.48}{7}\\ k &= 0.0494 \end{aligned}$$

(d)

$$\begin{aligned} \lg\left[1.25\left(10^{3.48}\right)\right] &= 3.58\\ t &= 6.7 \end{aligned}$$

Solution 2

(a)

$$\begin{aligned} n &= n_0(2)^{kt}\\ \ln n &= \ln(n_0) + (k\ln 2)t \end{aligned}$$ Draw a straight line of $\ln n$ against $t$.

Best fit line

(b)

$$\begin{aligned} \ln n_0 &= 8.00\\ n_0 &= e^8\\ &= 2980.957\\ &= 2980 \end{aligned}$$

(c)

$$\begin{aligned} k\ln 2 &= \frac{8.35 - 8.0}{10}\\ k &= 0.050494\\ &= 0.0505 \end{aligned}$$

(d)

$$\begin{aligned} \ln\left[1.25\left(e^8\right)\right] &= 8.223\\ t &= 6.3 \end{aligned}$$

(a)

$$\begin{aligned} x + py &= qxy\\ 1 + \frac{py}{x} &= qy\\ qy &= 1 + \frac{py}{x}\\ y &= \left(\frac{p}{q}\right)\frac{y}{x} + \left(\frac{1}{q}\right) \end{aligned}$$ Vertical intercept $= \dfrac{1}{q}$ $$q = \frac{1}{\text{Vertical intercept}}$$ Gradient $= \dfrac{p}{q}$ $$p = q(\text{Gradient})$$

(b)

$$\lg(y - 5) = m\lg x + c$$ When $x = 10$ & $y = 15$; $\lg x = 1$, $\lg(y - 5) = 1$.

When $x = 100$ & $y = 1005$; $\lg x = 2$, $\lg(y - 5) = 3$. $$\begin{aligned} \text{Gradient, } m &= \frac{3 - 1}{2 - 1} = 2\\ \lg(y - 5) &= 2\lg x + c\\ 1 &= 2(1) + c\\ c &= -1\\ \lg(y - 5) &= 2\lg x - 1 \end{aligned}$$ When $y = 50$, $$\begin{aligned} \lg 45 &= 2\lg x - 1\\ \lg x &= \frac{\lg 45 + 1}{2}\\ x &= 10^{\frac{\lg 45 + 1}{2}} = 21.2 \end{aligned}$$

(a)

$$\begin{aligned} \text{gradient of straight line: } \frac{3 - 1}{4 - 2} &= 1 \end{aligned}$$ equation: $$\begin{aligned} Y - 1 &= 1(X - 2)\\ Y &= X - 1\\ \frac{x}{y} &= \sqrt{x} - 1\\ y &= \frac{x}{\sqrt{x} - 1} \end{aligned}$$

(b)

(i) $$\begin{aligned} y &= \frac{b^x}{10^a} \implies \lg y = \lg\left(\frac{b^x}{10^a}\right)\\ \lg y &= \lg b^x - \lg 10^a\\ \lg y &= x\lg b - a \end{aligned}$$

(ii) $$\begin{aligned} \lg y &= x\lg b - a\\ -a &= 1.5\\ a &= -1.5\\ \lg b &= 0.3\\ b &= 2.00 \end{aligned}$$

(iii) $$\begin{aligned} (10b)^x &= 10^{a+1}\\ 10^x b^x &= 10^a \times 10\\ \frac{b^x}{10^a} &= \frac{10}{10^x}\\ y &= 10^{1-x}\\ \lg y &= \lg\left(10^{1-x}\right)\\ \lg y &= 1 - x \end{aligned}$$ Draw the line $\lg y = 1 - x$ and find the $x$-coordinate of the point of intersection.

OR

$$\begin{aligned} (10b)^x &= 10^{a+1}\\ x\lg(10b) &= (a + 1)\lg(10)\\ x\lg(10) + x\lg(b) &= a + 1\\ x\lg(b) - a &= x + 1\\ \lg y &= x + 1 \end{aligned}$$

(a)

(b)

(i) The incorrect reading of $I$ is $4.22$.

From the graph, the correct $\lg I$ reading is $0.95$. $$\begin{aligned} 0.95 &= \lg I\\ I &= 10^{0.95}\\ I &= 8.91 \ (\text{3 sf}) \end{aligned}$$

(ii) $$\begin{aligned} I &= Ad^n\\ \lg I &= n\lg d + \lg A \end{aligned}$$ From the graph $$\begin{aligned} \lg A &= 3.6\\ A &= 10^{3.6}\\ &\approx 3981.071\\ &= 3980 \ (\text{3sf}) \end{aligned}$$ $$\begin{aligned} \text{Gradient} &= \frac{0.5 - 3.6}{1.52 - 0}\\ &\approx -2.0394\\ &= -2.04 \ (\text{3sf})\\ n &= -2.04 \end{aligned}$$

(c)

From the table, $$\frac{I_1}{I_2} = \frac{35}{141} = 0.248 \ (\text{to 3 sf})$$ When the distance doubled, the intensity decrease to $0.248$ (one-quarter) of its previous value.

(a)

(i) $$\begin{aligned} V &= ab^{2t}\\ \lg V &= \lg a + 2t\lg b \end{aligned}$$ Plot $\lg V$ against $t$

$\lg a =$ vertical intercept

$2\lg b =$ gradient

Alternative:

(ii)

$$\lg a = 0.17\ [0.16 - 0.18] \qquad 2\lg b = \frac{1 - 0.24}{5.3 - 0.5} = \frac{19}{120} = 0.158\ [0.14 \text{ to } 0.18]$$ $$a = 10^{0.17} \qquad\qquad\qquad\qquad \lg b = \frac{19}{240}$$ $$\phantom{a} = 1.48\ [1.445 \text{ to } 1.513] \qquad\qquad b = 1.20\ [1.17 \text{ to } 1.23]$$

(iii) From the graph, at $t = 6$ $$\begin{aligned} \lg V &= 1.12\\ V &= 13.2\ [13.2 \text{ to } 13.5] \end{aligned}$$ When $t = 6$ hours, there are 13.2 thousands of views

(b)

$$\begin{aligned} Y &= xy\\ X &= x^2\\ m &= \frac{9 - 1}{1 - 5} = -2\\ Y - 9 &= -2(X - 1)\\ Y &= -2X + 11\\ xy &= -2x^2 + 11\\ y &= \frac{-2x^2 + 11}{x}\\ &= -2x + \frac{11}{x} \end{aligned}$$