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.

*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.

*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.

*Correct!*

*True**False*Note that ${e}^{c}=x\Rightarrow lnx=c\phantom{\rule{0.3em}{0ex}}.$*True**False*Note that ${log}_{10}x=c\Rightarrow 1{0}^{c}=x\phantom{\rule{0.3em}{0ex}}.$

Which of the following solve this equation? Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is correct!*

then ${e}^{4x-2x}={e}^{2x}=0.6\phantom{\rule{0.3em}{0ex}},$ so we take the natural log of both sides

$ln\left({e}^{2x}\right)=2x=ln0.6\Rightarrow x=\frac{ln0.6}{2}=-0.0766\phantom{\rule{0.3em}{0ex}}.$

*Choice (c) is incorrect*

*Choice (d) is incorrect*

There may be more than one correct answer. (Zero or more options can be correct)

*There is at least one mistake.*

For example, choice (a) should be False.

*There is at least one mistake.*

For example, choice (b) should be True.

*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.

*Correct!*

*False*Recall $ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{0.3em}{0ex}}.$*True*$\sqrt{3x+1}={\left(3x+1\right)}^{\frac{1}{2}}\text{and}\phantom{\rule{1em}{0ex}}ln{a}^{b}=blna\phantom{\rule{0.3em}{0ex}}.$*True*$ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}ln{a}^{b}=blna\phantom{\rule{0.3em}{0ex}}.$*False*Recall $ln\left(\frac{a}{b}\right)=lna-lnb\phantom{\rule{0.3em}{0ex}}.$

Which of the following gives the population, $P$ (in millions), $t$ years after 1980? Exactly one option must be correct)

*Choice (a) is incorrect*

*Choice (b) is incorrect*

*Choice (c) is incorrect*

*Choice (d) is correct!*

$P={P}_{0}^{}{e}^{kt}$ 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 ${e}^{10k}=1.25\Rightarrow ln\left({e}^{10k}\right)=10k=ln1.25$

So $k=\frac{ln1.25}{10}=0.022$ and $P\left(t\right)=2{e}^{0.022t}\phantom{\rule{0.3em}{0ex}}.$