Natural logarithms and exponential models

What you’ll learn

• Solve equations involving e and natural logarithms.
• Interpret growth and decay models.
• Linearise power and exponential relationships.

Understand the idea

The functions $e^x$ and $\ln x$ are inverses, so $\ln(e^x)=x$ and $e^{\ln x}=x$ for $x>0$. The exponential graph is always positive; the logarithm graph exists only for positive inputs. In equations, isolate the exponential or logarithm before applying its inverse.

A model $N=N_0e^{kt}$ has initial value $N_0$ and a constant proportional rate of change $k$. Positive $k$ means growth, negative $k$ decay. A real process may eventually outgrow this model if resources, background levels or other conditions change; interpret predictions in the model's stated time range.

Logarithms turn certain curves into straight lines. For $y=ax^n$, plotting $\ln y$ against $\ln x$ gives gradient $n$ and intercept $\ln a$. For $y=kb^x$, plotting $\ln y$ against $x$ gives gradient $\ln b$ and intercept $\ln k$. Recover the original constants by exponentiating the intercept or gradient as appropriate.

Key formulae

$$\ln(e^x)=x,\qquad e^{\ln x}=x\quad(x>0)$$

$$N=N_0e^{kt},\qquad \frac{dN}{dt}=kN$$

$$y=ax^n\Rightarrow\ln y=\ln a+n\ln x$$

$$y=kb^x\Rightarrow\ln y=\ln k+x\ln b$$

Worked example

A population model is $N=240e^{0.08t}$. Find the time for the population to double, to 2 decimal places.

1. Set $N=480$: $480=240e^{0.08t}$.
2. Divide by $240$ and take natural logs: $\ln2=0.08t$.
3. Compute $t=\ln2/0.08=8.6643\ldots$.

Answer: $t=8.66$ time units.