 ## H1011 Quizzes

Quiz 3: Exponential and logarithmic functions
Question 1 Questions
Write $3sinx+4cosx$ in the form $Rsin\left(x+\alpha \right)$.
 a) $\sqrt{7}sin\left(x+0.64\right)$ b) $\sqrt{7}sin\left(x+0.93\right)$ c) $5sin\left(x+0.64\right)$ d) $5sin\left(x+0.93\right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$Rcos\alpha =3$, $Rsin\alpha =4$, $R=\sqrt{{3}^{2}+{4}^{2}}=5$, $\alpha ={tan}^{-1}4∕3=0.93$
Write $\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx$ in the form $Rsin\left(x+\alpha \right)$.
 a) $sin\left(x+\frac{\pi }{6}\right)$ b) $sin\left(x-\frac{\pi }{6}\right)$ c) $sin\left(x+\frac{5\pi }{6}\right)$ d) $sin\left(x-\frac{5\pi }{6}\right)$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
$Rcos\alpha =-\frac{\sqrt{3}}{2}$, $Rsin\alpha =\frac{1}{2}$, $R=\sqrt{\frac{1}{3}+\frac{3}{4}}=1,\phantom{\rule{1em}{0ex}}tan\alpha =-\frac{1}{\sqrt{3}}$ and $\alpha$ is in the 2nd quadrant.
Choice (d) is incorrect
Which of the following graphs indicates that $y$ is proportional to ${x}^{2}$.
 a) b) c) d) Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
Choice (d) is incorrect
The experimental data $\begin{array}{cccccc}\hfill t\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 0.9\hfill & \hfill 3.6\hfill & \hfill 8.1\hfill & \hfill 14.4\hfill & \hfill 22.5\hfill \end{array}$ appears to show that $y$ is proportional to ${t}^{2}$. Find the relationship between $y$ and $t$.
 a) $y=0.9{t}^{2}$ b) $y=2.7{t}^{2}$ c) $y=0.9t$ d) $y=2.7t$

Choice (a) is correct!
Scaling the data gives $\begin{array}{cccccc}\hfill {t}^{2}=T\hfill & \hfill 1\hfill & \hfill 4\hfill & \hfill 9\hfill & \hfill 16\hfill & \hfill 25\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 0.9\hfill & \hfill 3.6\hfill & \hfill 8.1\hfill & \hfill 14.4\hfill & \hfill 22.5\hfill \end{array}$ A scatter plot of $y$ against $T$ is a straight line through (0,0) with slope $\frac{3.6-0.9}{4-1}=0.9$.
$\text{therefore,}y=0.9T=0.9{t}^{2}$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Given the experimental data $\begin{array}{cccccc}\hfill t\hfill & \hfill 1.1\hfill & \hfill 1.6\hfill & \hfill 2.1\hfill & \hfill 2.6\hfill & \hfill 3.1\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 2.9\hfill & \hfill 4.4\hfill & \hfill 5.9\hfill & \hfill 7.4\hfill & \hfill 8.9\hfill \end{array}$ which of the following statements is correct ?
 a) $y$ is proportional to $t$. b) $y$ is proportional to ${t}^{2}$. c) $y$ is a linear function of $t$. d) There is no recognisable relationship between $y$ and $t$.

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
The equation of the line satisfying the data is $y=3t-0.4$.
Choice (d) is incorrect
Solve $3{e}^{x}=5$.
 a) $x=\frac{ln5}{3}$ b) $x=\frac{ln5}{ln3}$ c) $x=ln5+ln\left(-3\right)$ d) $x=ln5-ln3$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is correct!
$\begin{array}{rcll}3{e}^{x}=5⇒{e}^{x}& =& \frac{5}{3}& \text{}\\ ln{e}^{x}& =& ln\left(\frac{5}{3}\right)& \text{}\\ x& =& ln5-ln3.& \text{}\end{array}$
Solve $\frac{1}{3}ln8{x}^{3}=2.5$ .
 a) $x=\frac{{e}^{2.5}}{2}$ b) $x=\sqrt{7.5-ln8}$ c) $x=3{e}^{7.5-ln8}$ d) $x=\frac{1}{3}{e}^{7.5∕8}$

Choice (a) is correct!
$\begin{array}{rcll}\frac{1}{3}ln8{x}^{3}& =& 2.5& \text{}\\ ln{\left[{\left(2x\right)}^{3}\right]}^{1∕3}& =& 2.5& \text{}\\ ln2x=2.5& & & \text{}\\ 2x={e}^{2.5}& & & \text{}\\ x=\frac{{e}^{2.5}}{2}.& & & \text{}\end{array}$
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
Simplify ${e}^{2ln\sqrt{{x}^{3}}}$.
 a) $2{x}^{3}$ b) ${e}^{2}+{x}^{3∕2}$ c) ${x}^{3}$ d) $2\sqrt{{x}^{3}}$

Choice (a) is incorrect
Choice (b) is incorrect
Choice (c) is correct!
${e}^{2ln\sqrt{{x}^{3}}}={e}^{2ln{x}^{3∕2}}={e}^{ln{\left({x}^{3∕2}\right)}^{2}}={e}^{ln{x}^{3}}={x}^{3}$.
Choice (d) is incorrect
It is expected that the set of experimental data $\begin{array}{ccccccc}\hfill x\hfill & \hfill 0.8\hfill & \hfill 1.4\hfill & \hfill 2.0\hfill & \hfill 2.6\hfill & \hfill 3.2\hfill & \hfill 3.8\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 1.0\hfill & \hfill 5.5\hfill & \hfill 16.0\hfill & \hfill 35.1\hfill & \hfill 65.5\hfill & \hfill 109.7\hfill \end{array}$ follows a power law. Find the best approximation for the given data.
 a) $y=1{0}^{-2}{e}^{7.5x}$ b) $y=2×1{0}^{-3}{e}^{7.5x}$ c) $y=1.1{x}^{1∕3}$ d) $y=2{x}^{3}$

Choice (a) is incorrect
This equation does not follow a power law.
Choice (b) is incorrect
This equation does not follow a power law.
Choice (c) is incorrect
Choice (d) is correct!
We do a log-log transformation and to 2 decimal places we find $\begin{array}{ccccccc}\hfill lnx=X\hfill & \hfill -0.22\hfill & \hfill 0.34\hfill & \hfill 0.69\hfill & \hfill 0.96\hfill & \hfill 1.16\hfill & \hfill 1.34\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill lny=Y\hfill & \hfill 0\hfill & \hfill 1.70\hfill & \hfill 2.77\hfill & \hfill 3.56\hfill & \hfill 4.18\hfill & \hfill 4.70\hfill \end{array}$ A scatterplot shows that the relationship between $X$ and $Y$ is linear.You should draw the scatterplot to verify this. So putting $Y=mX+b$ we find $m=3$ thus $Y=3X+b$. Now $0=-0.66+b$ therefore $y={e}^{0.66}{x}^{3}\approx 2{x}^{3}$.
It is expected that the set of experimental data $\begin{array}{cccccc}\hfill x\hfill & \hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill & \hfill 5\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill y\hfill & \hfill 3.0\hfill & \hfill 8.5\hfill & \hfill 15.6\hfill & \hfill 24.0\hfill & \hfill 33.5\hfill \end{array}$ follows a power law. Find the best approximation for the given data.
 a) $y=1.1{e}^{x}$ b) $y=1.2{e}^{0.9x}$ c) $y=3{x}^{3∕2}$ d) $y=3{x}^{2∕3}$

Choice (a) is incorrect
This equation does not follow a power law.
Choice (b) is incorrect
This equation does not follow a power law.
Choice (c) is correct!
We do a log-log transformation and to 2 decimal places we find $\begin{array}{cccccc}\hfill lnx=X\hfill & \hfill 0\hfill & \hfill 0.69\hfill & \hfill 1.10\hfill & \hfill 1.39\hfill & \hfill 1.61\hfill \\ ̲& ̲& ̲& ̲& ̲& ̲\\ \hfill lny=Y\hfill & \hfill 1.10\hfill & \hfill 2.14\hfill & \hfill 2.75\hfill & \hfill 3.18\hfill & \hfill 3.51\hfill \end{array}$ A scatterplot shows that the relationship between $X$ and $Y$ is linear. You should draw the scatterplot to verify this. So letting $Y=mX+b$ we find $m=1.5$ and $b=1.1$. Thus $y={e}^{1.1}{x}^{1.5}\approx 3{x}^{1.5}=3{x}^{3∕2}$.
Choice (d) is incorrect