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In this graph, I am asked to find out that the equation satisfies the straight line on the graph. Therefore I have evaluated the slope of the graph by the two coordinates, (3,10) and (300, 100 ) which are the two ends of the straight line.

The slope is, $\frac{10}{33}$.

The answer given is: $$y = 6\sqrt{x}$$.

My question is , cant I use the slope to find the equation? If not then what is the strategy to find the equation?

  • 3
    $\begingroup$ That's exactly what you can do. Bear in mind that the equation of a line in the log-log plane is not $y=m\cdot x + b$, but rather $\log y = m \cdot \log x + b$. Work from that and you can solve for $y$ as a function of $x$ $\endgroup$
    – Richard P
    Oct 21 '15 at 22:24

You have an equation of the form $$ \ln y = a\ln x + b $$ and the two points you have give $$\begin{align} \ln 10 &= a\ln 3 + b\\ \ln 100 &= a\ln 300 + b. \end{align}$$ In particular, solving the linear system for $a,b$ you get $$\begin{align} a &= \frac{\ln 10}{\ln 100} = \frac{1}{2}\\ b &= \ln 10 - \frac{1}{2}\ln 3 = \frac{1}{2}\ln \frac{100}{3}. \end{align}$$

Plugging it back, $$ \ln y = \frac{1}{2}\ln x + \frac{1}{2}\ln \frac{100}{3} $$ or, exponentiating both sides, $$ y = \frac{10}{\sqrt{3}}\sqrt{x} \simeq 5.8\sqrt{x} $$


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