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Log-log plots problems (not HW, but highly recommended).

For each of the data sets below, graph these points on either semi-log or log-log paper and determine the function which best models these data points from the choices below. 
\vskip 5pt

1.)         (1, 10),  (8, 40),  (32, 100),  (8000, 4000)~~~~ANS:  $y = 10t^\frac{2}{3}$

%%2.) (1, 10000),  (2, 3900), (6, 620), (7, 290)


2.) (1, 10000),                         (5, 400), (15, 50), (73, 2)~~~~ANS:  $y = 10000t^{-2}$


%%4.)  (0, 1), (1.4, 3), (4.4, 30), (8, 480) 

%%5.)  (0,  100),  (2, 45 ), (3.2, 7), (4, 2)

3.)  (1, 1), (60, 4), (200, 6), (3200, 15), (8000, 20)~~~~ANS:  $y = t^\frac{1}{3}$

4.)  (1, 1000),                        (5, 200), (20, 50), (515, 2)~~~~ANS:  $y = 1000t^{-1}$

%%8.)  (0, 10), (0.6, 40), (1.8, 605), (2, 1000)

5.) (1, 100),               (35, 600), (400, 2000), (8100, 9000)~~~~ANS:  $y = 100t^\frac{1}{2}$

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\vskip 5pt

\textbf{A)} $y =  0$ \hfill
\textbf{B)}   $y =  t^{\frac{1}{3}}$ \hfill
\textbf{C)}   $y =  t^{\frac{1}{2}}$ \hfill
\textbf{D)}   $y =  t^{\frac{2}{3}}$ \hfill
\textbf{E)} $y =  10^t$ \hfill
\textbf{F)} $y =  t^{\frac{3}{2}}$ \hfill
\textbf{G)} $y =  t^{2}$\\ \\
\textbf{H)} $y =  1$ \hfill
\textbf{I)}   $y =  t^{-\frac{1}{3}}$ \hfill
\textbf{J)}   $y =  t^{-\frac{1}{2}}$ \hfill
\textbf{K)}   $y =  t^{-\frac{2}{3}}$ \hfill
\textbf{L)} $y =  t^{-1}$ \hfill
\textbf{M)} $y =  t^{-\frac{3}{2}}$ \hfill
\textbf{N)} $y =  t^{-2}$\\ \vskip 2pt

\textbf{0)} $y =  10$ \hfill
\textbf{P)}   $y =  10t^{\frac{1}{3}}$ \hfill
\textbf{Q)}   $y =  10t^{\frac{1}{2}}$ \hfill
\textbf{R)}   $y =  10t^{\frac{2}{3}}$ \hfill
\textbf{S)} $y =  10t$ \hfill
\textbf{T)} $y =  10t^{\frac{3}{2}}$ \hfill
\textbf{U)} $y =  10t^{2}$\\ \\
\textbf{V)}   $y =  10t^{-\frac{1}{3}}$ \hfill
\textbf{W)}   $y =  10t^{-\frac{1}{2}}$ \hfill
\textbf{X)}   $y =  10t^{-\frac{2}{3}}$ \hfill
\textbf{Y)} $y =  10t^{-1}$ \hfill
\textbf{Z)} $y =  10t^{-\frac{3}{2}}$ \hfill
\textbf{ZZ)} $y =  10t^{-2}$\\ \vskip 4pt

\textbf{a)} $y =  100$ \hfill
\textbf{b)}   $y =  100t^{\frac{1}{3}}$ \hfill
\textbf{c)}   $y =  100t^{\frac{1}{2}}$ \hfill
\textbf{d)}   $y =  100t^{\frac{2}{3}}$ \hfill
\textbf{e)} $y =  100t$ \hfill
\textbf{f)} $y =  100t^{\frac{3}{2}}$ \hfill
\textbf{g)} $y =  100t^{2}$\\ \\
\textbf{h)}   $y =  100t^{-\frac{1}{3}}$ \hfill
\textbf{i)}   $y =  100t^{-\frac{1}{2}}$ \hfill
\textbf{j)}   $y =  100t^{-\frac{2}{3}}$ \hfill
\textbf{k)} $y =  100t^{-1}$ \hfill
\textbf{l)} $y =  100t^{-\frac{3}{2}}$ \hfill
\textbf{m)} $y =  100t^{-2}$\\ \vskip 4pt

\textbf{n)}   $y =  1000t^{\frac{1}{3}}$ \hfill
\textbf{o)}   $y =  1000t^{\frac{1}{2}}$ \hfill
\textbf{p)}   $y =  1000t^{\frac{2}{3}}$ \hfill
\textbf{q)} $y =  1000t$ \hfill
\textbf{r)} $y =  1000t^{\frac{3}{2}}$ \hfill
\textbf{s)} $y =  1000t^{2}$\\ \\
\textbf{t)}   $y =  1000t^{-\frac{1}{3}}$ \hfill
\textbf{u)}   $y =  1000t^{-\frac{1}{2}}$ \hfill
\textbf{v)}   $y =  1000t^{-\frac{2}{3}}$ \hfill
\textbf{x)} $y =  1000t^{-1}$ \hfill
\textbf{y)} $y =  1000t^{-\frac{3}{2}}$ \hfill
\textbf{z)} $y =  1000t^{-2}$\\ \vskip 4pt



\textbf{n)}   $y =  10000t^{\frac{1}{3}}$ \hfill
\textbf{o)}   $y =  10000t^{\frac{1}{2}}$ \hfill
\textbf{p)}   $y =  10000t^{\frac{2}{3}}$ \hfill
\textbf{q)} $y =  10000t$ \hfill
\textbf{r)} $y =  10000t^{\frac{3}{2}}$ \hfill
\textbf{s)} $y =  10000t^{2}$\\ \\
\textbf{t)}   $y =  10000t^{-\frac{1}{3}}$ \hfill
\textbf{u)}   $y =  10000t^{-\frac{1}{2}}$ \hfill
\textbf{v)}   $y =  10000t^{-\frac{2}{3}}$ \hfill
\textbf{x)} $y =  10000t^{-1}$ \hfill
\textbf{y)} $y =  10000t^{-\frac{3}{2}}$ \hfill
\textbf{z)} $y =  10000t^{-2}$\\ \vskip 4pt 


 
\textbf{B)}   $y =  10^{\frac{t}{3}}$ \hfill
\textbf{C)}   $y =  10^{\frac{t}{2}}$ \hfill
\textbf{D)}   $y =  10^{\frac{2t}{3}}$ \hfill
\textbf{E)} $y =  10^(10^t$ \hfill
\textbf{F)} $y =  10^{\frac{3t}{2}}$ \hfill
\textbf{G)} $y =  10^{2t}$\\ \\
\textbf{I)}   $y =  10^{-\frac{t}{3}}$ \hfill
\textbf{J)}   $y =  10^{-\frac{t}{2}}$ \hfill
\textbf{K)}   $y =  10^{-\frac{2t}{3}}$ \hfill
\textbf{L)} $y =  10^{-t}$ \hfill
\textbf{M)} $y =  10^{-\frac{3t}{2}}$ \hfill
\textbf{N)} $y =  10^{-2t}$\\ \vskip 4pt


\textbf{P)}   $y =  10(10^{\frac{t}{3}})$ \hfill
\textbf{Q)}   $y =  10(10^{\frac{t}{2}})$ \hfill
\textbf{R)}   $y =  10(10^{\frac{2t}{3}})$ \hfill
\textbf{S)} $y =  10(10^t)$ \hfill
\textbf{T)} $y =  10(10^{\frac{3t}{2}})$ \hfill
\textbf{U)} $y =  10(10^{2t})$\\ \\
\textbf{W)}   $y =  10(10^{-\frac{t}{2}})$ \hfill
\textbf{X)}   $y =  10(10^{-\frac{2t}{3}})$ \hfill
\textbf{Y)} $y =  10(10^{-t})$ \hfill
\textbf{Z)} $y =  10(10^{-\frac{3t}{2}})$ \hfill
\textbf{ZZ)} $y =  10(10^{-2t})$\\ \vskip 4pt


\textbf{c)}   $y =  100(10^{\frac{t}{2}})$ \hfill
\textbf{d)}   $y =  100(10^{\frac{2t}{3}})$ \hfill
\textbf{e)} $y =  100(10^t)$ \hfill
\textbf{f)} $y =  100(10^{\frac{3t}{2}})$ \hfill
\textbf{g)} $y =  100(10^{2t})$\\ \\
\textbf{i)}   $y =  100(10^{-\frac{t}{2}})$ \hfill
\textbf{j)}   $y =  100(10^{-\frac{2t}{3}})$ \hfill
\textbf{k)} $y =  100(10^{-t})$ \hfill
\textbf{l)} $y =  100(10^{-\frac{3t}{2}})$ \hfill
\textbf{m)} $y =  100(10^{-2t})$\\ \vskip 4pt


%%\textbf{o)}   $y =  1000(10^{\frac{t}{2}})$ \hfill
%%\textbf{p)}   $y =  1000(10^{\frac{2t}{3}})$ \hfill
%%\textbf{q)} $y =  1000(10^t)$ \hfill
%%\textbf{r)} $y =  1000(10^{\frac{3t}{2}})$ \hfill
%%\textbf{s)} $y =  1000(10^{2t})$\\ \\
%%\textbf{u)}   $y =  1000(10^{-\frac{t}{2}})$ \hfill
%%\textbf{v)}   $y =  1000(10^{-\frac{2t}{3}})$ \hfill
%%\textbf{x)} $y =  1000(10^{-t})$ \hfill
%%\textbf{y)} $y =  1000(10^{-\frac{3t}{2}})$ \hfill
%%\textbf{z)} $y =  1000(10^{-2t})$\\ \vskip 4pt


\eject

When to use log-log paper:\\

Suppose you suspect your data points satisfy polynomial growth of the form $y = At^m$ for some constants $A$ and $m$.  
\\

$y = At^m$

$log(y) = log(At^m)$

$log(y) = log(A) + m log(t)$.~~   Let $z = log(y)$ and $x = log(t)$.  Then  

$z = log(A) + mx$.  
\\

$z = mx + log(A)$.  ~~That is we have the equation of a line where slope = $m$ ~ and $z$-intercept $= log(A)$.
\\

If $z = mx + b$, then $log(A) = b$.  ~~ Hence $A = 10^{log(A)} = 10^b$.\\


Hence to determine the constants $A$ and $m$ in $y = At^m$, graph $(t, y)$ on log-log paper (note this is the same as taking $z = log(y)$ and $x = log(t)$), and determine equation of best fit line,  $z = mx + b$.  Then $y = 10^b t^m$.
\\

However if the data points do not satisfy a best fit line, then the data points do NOT satisfy polynomial growth of the form $y = At^m$\\


\hrule
\vskip 6pt

When to use semi-log paper:\\

Suppose you suspect your data points satisfy exponential growth of the form $y = Ac^t$ for some constants $A$ and $c$.  
\\

$y = Ac^t$

$log(y) = log(Ac^t)$

$log(y) = log(A) + t log(c)$.~~   Let $z = log(y)$.  Then  

$z = log(A) + t log(c)$.  
\\

$z = [log(c)] t + log(A)$.  ~~I.e. we have the equation of a line where slope = $log(c)$ ~ and $z$-intercept $= log(A)$.
\\

If $z = mt + b$, then (i) $log(A) = b$.  ~ Hence $A = 10^{log(A)} = 10^b$.
~~~ (ii) $log(c) = m$.  ~Hence $c = 10^m$.
\\


Hence to determine the constants $A$ and $c$ in $y = Ac^t$, graph $(t, y)$ on semi-log paper (note this is the same as taking $z = log(y)$), and determine equation of best fit line,  $z = mx + b$.  Then $y = 10^b (10^m)^t$.  I.e., $y = 10^b (10^{mt})$
\\

However if the data points do not satisfy a best fit line, then the data points do NOT satisfy polynomial growth of the form $y = Ac^t$








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