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I'm just going through some further maths units as I prepare for a PHD in chemical engineering. I'm finding the thought processes to be invaluable in my problem solving skills.

However, I recently came unstuck on this question....and while I know I can't be helped here due to the requirement to construct a graph, I wondered what would be the right way a mathematician would approach this? Show graphically that the following figures support the belief that a relationship of the form =^ exists and use your graph to find approximate values for the constants and .

5 8 12 14 20 30 22.4 45.3 83.1 104.8 178.9 328.6

Thank you

New Zealand's finest
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1 Answers1

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If the relationship $y=ax^k$ holds, then taking logs of both sides yields $$ \log y = \log a + k\log x. $$ This means that plotting $\log y$ against $\log x$ should give a straight line with slope $k$ and intercept $\log a$.

grand_chat
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  • Thank you for this help. So all I need to do is plot a graph with log y and another with log a + k log x. Will the slope be the linkage? I will use desmos. – New Zealand's finest May 09 '16 at 13:13
  • Make a new data set with $\log x$ as the independent variable and $\log y$ as the dependent variable. Plot these pairs of variables -- $\log x$ on the horizontal axis, and $\log y$ on the vertical axis (instead of the original $x$ and $y$). These pairs should exhibit a straight line relationship, whereas the original pairs will not. The slope of the line will be $k$ and the intercept will be $\log a$. – grand_chat May 09 '16 at 21:07
  • What would be the quickest way to compute log x and log y as values? My methodology is lacking.

    So logy is not being changed. Logx is the independent variable and so changes in terms of value as it is the variable being measured.

     – New Zealand's finest May 10 '16 at 19:47
  • So for example, for the first value, I would just take the log of 5?

    Which would be 0.699

     – New Zealand's finest May 10 '16 at 19:54
  • I have achieved this now! – New Zealand's finest May 10 '16 at 21:38
  • Thanks for your assistance grand_chat! – New Zealand's finest May 10 '16 at 21:38
  • Great. I assume you used a calculator to compute log x and log y? (I don't get 0.699 as log of 5.) – grand_chat May 10 '16 at 21:40