1.06h - Linearising power and exponential models

You already know

For positive arguments, $\ln(uv)=\ln u+\ln v$ and $\ln(x^n)=n\ln x$. The inverse of $\ln$ is the exponential: $\ln a=c$ means $a=e^c$.

Linearisation

Linearisation transforms a relationship into straight-line form. For a power model, use $X=\ln x$ and $Y=\ln y$, with $x,y>0$. The transformed points lie on a straight line if the power model holds exactly; experimental scatter normally makes the line approximate.

Power-model transformation

For $y=ax^n$, where $a>0$, $x>0$ and $n$ is real,

Thus gradient $m=n$, intercept $c=\ln a$, and $a=e^c$. Using base-ten logs instead gives $a=10^c$, with the same gradient $n$.

Not in the formula sheet: learn it.

Worked example 1: recovering a power model

Question. An approximate fitted line for these measurements is $Y=1.50X+1.39$, where $X=\ln x$ and $Y=\ln y$. Find the power model.

Plan. Compare the fitted line with $Y=nX+\ln a$, then undo the logarithm of the coefficient.

Why this step? The vertical coordinate represents $\ln y$, so its intercept represents $\ln a$, not $a$.

Answer. $y\approx4.01x^{1.50}$.

Check. Using the unrounded coefficient at $x=4$ gives $e^{1.39}4^{1.5}\approx32.1$, close to the measurement $31.7$.

Worked example 2: a negative power

Question. A different experiment gives $\ln y=-1.50\ln x+1.39$. Find the power model and its prediction at $x=4$.

Plan. Read the exponent from the gradient and exponentiate the intercept as before.

Why this step? A negative gradient changes the exponent, not the sign of the coefficient.

Answer. $y=e^{1.39}x^{-1.50}$, with $y(4)\approx0.502$.

Check. Multiplying $x$ by $4$ multiplies $y$ by $4^{-1.5}=1/8$; $e^{1.39}/8\approx0.502$.

Watch out

From $\ln y=n\ln x+c$, writing $y=cx^n$ is wrong. The coefficient is $e^c$; the line's intercept is its logarithm.

Exponential-model transformation

For $y=kb^x$, where $k>0$ and $b>0$,

A plot of $Y=\ln y$ against $x$ has gradient $m=\ln b$ and intercept $c=\ln k$, so $b=e^m$ and $k=e^c$. Equivalently, $y=ke^{rx}$ gives gradient $r$.

Not in the formula sheet: learn it.

Worked example 3: an exponential fit

Question. Experimental data give the approximate line $\ln y=0.183x+1.79$. Find $k$ and $b$ in $y=kb^x$.

Plan. Exponentiate both the gradient and intercept using base $e$.

Why this step? Increasing $x$ by $1$ adds $0.183$ to $\ln y$, which multiplies $y$ by $e^{0.183}$.

Answer. $y=e^{1.79}(e^{0.183})^x\approx5.99(1.201)^x$, retaining extra digits in the base for use.

Check. At $x=0$, $\ln y=1.79$ and $y=e^{1.79}$, as required.

Worked example 4: reading the log base

Question. A fitted line is $\log_{10}y=0.183x+1.79$. Find its exponential model.

Plan. Undo each base-ten logarithm with a power of $10$.

Why this step? $10^u$, rather than $e^u$, is the inverse of $\log_{10}u$.

Answer. $y=10^{1.79}(10^{0.183})^x$.

Check. At $x=1$, $\log_{10}y=1.973$; multiplying the two recovered factors gives $10^{1.973}$.

Watch out

For $\ln y=mx+c$, writing $y=e^cm^x$ uses the gradient as the base. The correct model is $y=e^c(e^m)^x=e^{c+mx}$.