AHL 2.10—Scaling large numbers, log-log graphs

Taking $\ln$ of $P=k{e}^{mt}$ gives $\ln (P)=mt+\ln (k)$. On a $\ln (P)$ vs $t$ graph, the gradient equals $m$ directly.

Using points $(0,3.912)$ and $(6,5.298)$: $m=\frac{5.298-3.912}{6-0}=\frac{1.386}{6}=0.231$

Since the gradient of $\ln (P)$ vs $t$ equals $m$ directly, we have $m\approx 0.231$.

The gradient of the $\ln (P)$ vs $t$ line equals $m$. We need $\frac{\Delta \ln P}{\Delta t}=\frac{5.298-3.912}{6-0}=\frac{1.386}{6}$.

$0.693\approx \ln 2$. This would require $\Delta \ln P=0.693\times 6=4.158$, but the actual change is only $1.386$. Eliminated.

This would give $\Delta \ln P=0.462\times 6=2.772$, which does not match the observed change of $1.386$. Eliminated.

This gives $\Delta \ln P=0.154\times 6=0.924\ne 1.386$. This might result from dividing $1.386$ by $9$ instead of $6$. Eliminated.

$m=\frac{1.386}{6}=0.231$. The correct answer is $m\approx 0.231$.

For $F=a{r}^b$, taking ${\log }_{10}$ of both sides gives ${\log }_{10}(F)=b{\log }_{10}(r)+{\log }_{10}(a)$. The gradient is $b$ and the $y$-intercept is ${\log }_{10}(a)$.

Gradient $=b=-2$ and $y$-intercept $={\log }_{10}(a)=4.8$.

Since ${\log }_{10}(a)=4.8$, we get $a=10^{4.8}$. The model is $F=10^{4.8}\cdot r^{-2}$.

On a log-log graph, a straight line with gradient $m$ and intercept $c$ gives $b=m=-2$ and $a=10^c=10^{4.8}$.

This treats the $y$-intercept $4.8$ as the value of $a$ directly, forgetting to apply the inverse ${\log }_{10}$. Since ${\log }_{10}(a)=4.8$, $a=10^{4.8}\ne 4.8$. Eliminated.

This swaps the roles of gradient and intercept, using the intercept as the exponent and the gradient as the base of $10$. Eliminated.

Using $e^{4.8}$ is only appropriate when the $y$-intercept is $\ln (a)$, i.e., on a $\ln$ semi-log graph. Here we have ${\log }_{10}$, so $a=10^{4.8}$. Eliminated.

The correct model is $F=10^{4.8}\cdot r^{-2}$.