Logarithm Graphs

lovely_nancy

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Sep 14, 2010
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could I get help on how to answer the following question please.

It is believed that the relationship between the variables x and y is of the form y = Ax^n. In an experiment the data
in the table are obtained.

x 3, 6, 10, 15, 20
y 10.4, 29.4, 63.2, 116.2, 178.19


Draw the graph of log10y against log10x.
Explain and justify how the shape of your graph enables you to decide
whether the relationship is indeed pf the form y = Ax^n.
Estimate the value of A and n.
 
Hello, lovely_nancy!

It is believed that the relationship between the variables x and y is of the form: y=Axn\displaystyle \text{It is believed that the relationship between the variables }x\text{ and }y\text{ is of the form: }\:y \:=\: Ax^n

In an experiment the following data are obtained.\displaystyle \text{In an experiment the following data are obtained.}

. . [1][2][3][4][5]x36101520y10.429.463.2116.2178.19\displaystyle \begin{array}{|c||c|c|c|c|c|} \hline & [1] & [2] & [3] & [4] & [5] \\ \hline\hline x & 3 & 6 & 10 & 15 & 20 \\ \hline y & 10.4 & 29.4 & 63.2 & 116.2 & 178.19 \\ \hline \end{array}


Draw the graph of log10y against log10x.\displaystyle \text{Draw the graph of }\log_{10}y\text{ against }\log_{10}x.

Explain and justify how the shape of your graph enables you to decide\displaystyle \text{Explain and justify how the shape of your graph enables you to decide}
. . whether the relationship is indeed if the form: y=Axn\displaystyle \text{whether the relationship is indeed if the form: }\:y \:=\: Ax^n

We have: y  =  Axn\displaystyle \text{We have: }\:y \;=\;Ax^n

Take logs:   log(y)  =  log(Axn)  =  log(A)+log(xn)log(y)  =  nlog(x)+log(A)\displaystyle \text{Take logs: }\;\log(y) \;=\;\log(Ax^n) \;=\;\log(A) \,+\, \log(x^n) \quad\Rightarrow\quad \log(y)\;=\;n\log(x) \,+\, \log(A)

We have an equation of the form:   Y  =  mX+b, a straight line\displaystyle \text{We have an equation of the form: }\;Y \;=\;mX + b,\:\text{ a straight line}


We have these data:   logx0.47710.77821.00001.17611.3010logy1.01701.46831.80072.06522.2509\displaystyle \text{We have these data: }\;\begin{array}{|c||c|c|c|c|c|} \hline \log x & 0.4771 & 0.7782 & 1.0000 & 1.1761 & 1.3010 \\ \hline \log y & 1.0170 & 1.4683 & 1.8007 & 2.0652 & 2.2509 \\ \hline \end{array}

If we plot these points, we find that they "line up" and form a straight line.\displaystyle \text{If we plot these points, we find that they "line up" and form a straight line.}

Therefore, the original function is an exponential: y=Axn\displaystyle \text{Therefore, the original function is an exponential: }\:y \:=\:Ax^n




Estimate the value of A and n.\displaystyle \text{Estimate the value of }A\text{ and }n.

We have the function: Axn=y\displaystyle \text{We have the function: }\:Ax^n \:=\:y

. . From [1], we have: A3n=10.4    (a)\displaystyle \text{From [1], we have: }\:A\cdot 3^n \:=\:10.4\;\;(a)

. . From [2], we have: A6n=29.4    (b)\displaystyle \text{From [2], we have: }\:A\cdot 6^n \:=\:29.4\;\;(b)

Divide (b) by (a): A6nA3n=29.410.4(63)n=29.410.42n=29.410.4\displaystyle \text{Divide (b) by (a): }\:\frac{A\cdot6^n}{A\cdot3^n} \:=\:\frac{29.4}{10.4} \quad\Rightarrow\quad \left(\frac{6}{3}\right)^n \:=\:\frac{29.4}{10.4} \quad\Rightarrow\quad 2^n \:=\:\frac{29.4}{10.4}

Take logs:   log(2n)=ln(29.410.4)nln(2)=ln(29.410.4)\displaystyle \text{Take logs: }\;\log\left(2^n\right) \:=\:\ln\left(\frac{29.4}{10.4}\right) \quad\Rightarrow\quad n\cdot\ln(2) \:=\:\ln\left(\frac{29.4}{10.4}\right)

. . . . n=ln(29.410.4)ln(2)  =  1.499232627n    1.5\displaystyle n \:=\:\dfrac{\ln(\frac{29.4}{10.4})}{\ln(2)} \;=\;1.499232627 \quad\Rightarrow\quad \boxed{n \;\approx\;1.5}

Substitute into (a):   A31.5=10.4A  =  10.431.5=2.001480933A    2\displaystyle \text{Substitute into (a): }\;A\,\cdot\,3^{1.5} \:=\:10.4 \quad\Rightarrow\quad A \;=\;\frac{10.4}{3^{1.5}} \,=\,2.001480933 \quad\Rightarrow\quad \boxed{A \;\approx\;2}


The function is:   y  =  2x32\displaystyle \text{The function is: }\;y \;=\;2x^{\frac{3}{2}}

 
lovely_nancy said:
could I get help on how to answer

I'm a bit confused, Nancy. Did you want help on how to answer, or did you want camera-ready copy ?

 
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