## MATH1111 Quizzes

The Logarithmic Function Quiz
Web resources available Questions

This quiz tests the work covered in Lecture 3 and corresponds to Section 1.4 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).

There are further web quizzes at Wiley. Choose section 4 from this page.

Be aware that it doesn’t seem to accept the written answers so you will have to check whether your answers are correct when they print the correct answer.

There are two sections of economics quiz site that would be useful, Logarithmic and Square Root Graphs and Logarithmic Functions.

The Mathematics Learning Centre has a booklet on Introduction to Exponents and Logarithms and tutors who can help you with the concepts.

There is a further useful site at http://www.ping.be/ ping1339/exp.htm which considers both logarithmic and exponential functions, as does http://www.themathpage.com/aPreCalc/logarithmic-exponential-functions.htm. There is a useful animation at http://www.analyzemath.com/logfunction/logfunction.html which shows how logarithmic functions change with different parameters.

Which of the following statements are correct? There may be more than one correct answer. (Zero or more options can be correct)
 a) ${e}^{2.35}=10.49⇒ln10.49=2.35\phantom{\rule{0.3em}{0ex}}.$ b) ${e}^{2.35}=10.49⇒ln2.35=10.49\phantom{\rule{0.3em}{0ex}}.$ c) ${log}_{10}17=1.23⇒1{0}^{1.23}=17\phantom{\rule{0.3em}{0ex}}.$ d) ${log}_{10}17=1.23⇒1{7}^{10}=1.23\phantom{\rule{0.3em}{0ex}}.$

There is at least one mistake.
For example, choice (a) should be True.
There is at least one mistake.
For example, choice (b) should be False.
Note that ${e}^{c}=x⇒lnx=c\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (c) should be True.
There is at least one mistake.
For example, choice (d) should be False.
Note that ${log}_{10}x=c⇒1{0}^{c}=x\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. True
2. False Note that ${e}^{c}=x⇒lnx=c\phantom{\rule{0.3em}{0ex}}.$
3. True
4. False Note that ${log}_{10}x=c⇒1{0}^{c}=x\phantom{\rule{0.3em}{0ex}}.$
Consider the equation  $3{e}^{2x}=5{e}^{4x}\phantom{\rule{0.3em}{0ex}}.$
Which of the following solve this equation? Exactly one option must be correct)
 a) ${e}^{2}=0.6\phantom{\rule{0.3em}{0ex}}.$ b) $x=-0.0766\phantom{\rule{0.3em}{0ex}}.$ c) $x=-0.3065\phantom{\rule{0.3em}{0ex}}.$ d) $x=1.0217\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, you have not used the index laws correctly.
Choice (b) is correct!
Rearrange the equation to become $\frac{{e}^{4x}}{{e}^{2x}}=\frac{3}{5}=0.6$
then ${e}^{4x-2x}={e}^{2x}=0.6\phantom{\rule{0.3em}{0ex}},$ so we take the natural log of both sides
$ln\left({e}^{2x}\right)=2x=ln0.6⇒x=\frac{ln0.6}{2}=-0.0766\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, you may have tried to solve ${e}^{0.5x}=\frac{3}{5}\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, you may have tried to solve ${e}^{2x}=\frac{5}{3}\phantom{\rule{0.3em}{0ex}}.$
Which of the following are written correctly as simpler logarithmic quantities?
There may be more than one correct answer. (Zero or more options can be correct)
 a) $ln\left(\frac{x+1}{x+2}\right)=ln\left(x+1\right)+ln\left(x=2\right)\phantom{\rule{0.3em}{0ex}}.$ b) $ln\sqrt{3x+1}=\frac{1}{2}ln\left(3x+1\right)\phantom{\rule{0.3em}{0ex}}.$ c) $ln\left(\frac{{x}^{2}\sqrt{x+1}}{\sqrt[3]{3x+4}}\right)=ln{x}^{2}+ln\sqrt{x+1}-ln\sqrt[3]{3x+4}\phantom{\rule{0.3em}{0ex}}.$ d) $ln\left(\frac{{x}^{2}\sqrt{x+1}}{\sqrt[3]{3x+4}}\right)=2lnx+\frac{1}{2}ln\left(x+1\right)+\frac{1}{3}ln\left(3x+4\right)\phantom{\rule{0.3em}{0ex}}.$

There is at least one mistake.
For example, choice (a) should be False.
Recall $ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (b) should be True.
$\sqrt{3x+1}={\left(3x+1\right)}^{\frac{1}{2}}\text{and}\phantom{\rule{1em}{0ex}}ln{a}^{b}=blna\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (c) should be True.
$ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}ln{a}^{b}=blna\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (d) should be False.
Recall $ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. False Recall $ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{0.3em}{0ex}}.$
2. True $\sqrt{3x+1}={\left(3x+1\right)}^{\frac{1}{2}}\text{and}\phantom{\rule{1em}{0ex}}ln{a}^{b}=blna\phantom{\rule{0.3em}{0ex}}.$
3. True $ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}ln{a}^{b}=blna\phantom{\rule{0.3em}{0ex}}.$
4. False Recall $ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{0.3em}{0ex}}.$
The population of a city increases at a rate which is proportional to the current population and was 2 million in 1980 and 2.5 million in 1990.
Which of the following gives the population, $P$ (in millions), $t$ years after 1980? Exactly one option must be correct)
 a) $P\left(t\right)=2.5{e}^{-0.022t}\phantom{\rule{0.3em}{0ex}}.$ b) $P\left(t\right)=2{e}^{-0.022t}\phantom{\rule{0.3em}{0ex}}.$ c) $P\left(t\right)=2.5{e}^{0.022t}\phantom{\rule{0.3em}{0ex}}.$ d) $P\left(t\right)=2{e}^{0.022t}\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, you seem to have confused the initial population and the population after 10 years.
Choice (b) is incorrect
Try again, you seem to have confused the initial population and the population after 10 years in your working.
Choice (c) is incorrect
Try again, you seem to have confused the initial population and the population after 10 years.
Choice (d) is correct!
Since $P$ increases at a rate which is proportional to the current population
$P={P}_{0}^{}{e}^{kt}$ where ${P}_{0}^{}=2$ million, the population in 1980.
We have that $2.5=2{e}^{10k}$ since the population is 2.5 million after 10 years.
Solving this for $k$ we have ${e}^{10k}=1.25⇒ln\left({e}^{10k}\right)=10k=ln1.25$
So $k=\frac{ln1.25}{10}=0.022$ and $P\left(t\right)=2{e}^{0.022t}\phantom{\rule{0.3em}{0ex}}.$