Problem — Exponent from a bilog graph

Problem

On doubly logarithmic paper, the line through the data $(x,y)$ has slope $+1{,}5$ and passes through the point $(1,4)$. Write the law $y(x)$, recalling that the slope in a bilogarithmic graph coincides with the exponent.

Solution

A power law $y = a\,x^n$ becomes a straight line on a doubly logarithmic scale: $\log y = \log a + n\,\log x$ where the slope of the line is the exponent $n$ and the intercept is $\log a$.

Exponent. The given slope is $1{,}5$, so: $n = 1{,}5$

Coefficient. The point $(1,4)$ means $x = 1,\ y = 4$. Substituting, since $1^n = 1$: $4 = a \cdot 1^{1{,}5} = a \quad\Rightarrow\quad a = 4$

Final law. $\ev{y = 4\,x^{1{,}5}}$

Links

Topics: Experimental method Skills: Using logarithmic scales · Reading graphs · Linearising data