Logarithm Graphs

[lovely_nancy]

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.

 

[soroban]

Hello, lovely_nancy!

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

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

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


\(\displaystyle \text{Draw the graph of }\log_{10}y\text{ against }\log_{10}x.\)

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

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

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

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


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

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

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

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

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

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

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

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

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

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

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


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

 

[mmm4444bot]

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 ?