### The Onion Method for Rearranging Equations

#### November 19, 2017

The other day in the office we were discussing methods of explaining how to change the subject of a formula, our different approaches and any alternatives we have. One difficulty that students can have is the order in which to rearrange and one of my colleagues demonstrated the onion method which I think might help some students. Below I outline how the method works with a few of examples.

## Example 1

If we take an example of the equation of motion below, where we want to change the subject to $$t$$.

$\require{enclose} v = u + at$ The onion method starts by drawing a circle around the new subject. $v = u + a\enclose{circle}{\:t\:}.$

Then we circle the term closest to it

$v = u + \enclose{circle}{\:a\:\enclose{circle}{\:t\:}\:}$

and continue with this until all the terms are circled

$v = \enclose{circle}{\:u + \enclose{circle}{\:a\:\enclose{circle}{\:t\:}\:}\:}.$

To rearrange we peel away the layers, so we start with the largest circle.

\begin{align} v &= \enclose{circle}{\:u + \enclose{circle}{\:a\:\enclose{circle}{\:t\:}\:}\:}. \\ v - u &= \enclose{circle}{\:u - u + \enclose{circle}{\:a\:\enclose{circle}{\:t\:}\:}\:}\\ v - u &= \enclose{circle}{\:a\:\enclose{circle}{\:t\:}\:} \end{align}

We continue with this approach

\begin{align} v - u &= \enclose{circle}{\:a\:\enclose{circle}{\:t\:}\:} \\ \frac{v - u}{a} &= \enclose{circle}{\:\frac{a\:\enclose{circle}{\:t\:}}{a}\:} \\ \frac{v - u}{a} &= \enclose{circle}{\:t\:} \end{align}

In this case we are now finished

$t = \frac{v - u}{a}$

but it works for more complicated examples too.

## Example 2

For a more complicated example we could use an equation that might model population growth and again change the subject to $$t$$.

$P = P_0e^{kt}$

As before we start by circling the new subject and then move to the next closest terms and so on outwards.

\begin{align} P &= P_0e^{k\enclose{circle}{\:t\:}} \\ P &= P_0e^{\enclose{circle}{\:k\enclose{circle}{\:t\:}\:}} \\ P &= P_0\enclose{circle}{\:\:\:e^{\enclose{circle}{\:k\enclose{circle}{\:t\:}\:}}\:} \\ P &= \enclose{circle}{\:P_0\enclose{circle}{\:\:\:e^{\enclose{circle}{\:k\enclose{circle}{\:t\:}\:}\:}\:}\:} \end{align}

Having added our layers we can peel them away.

\begin{align} P &= \enclose{circle}{\:P_0\enclose{circle}{\:\:\:e^{\enclose{circle}{\:k\enclose{circle}{\:t\:}\:}\:}\:}\:} \\ (\div P_0) & \:\:\:\:\:\:\:\: (\div P_0) \\ \frac{P}{P_0} &= \enclose{circle}{\:\:\:e^{\enclose{circle}{\:k\enclose{circle}{\:t\:}\:}\:}\:} \end{align}

Remembering we use $$\ln(x)$$ to cancel $$e^x$$ give us the next step.

\begin{align} \ln\left(\frac{P}{P_0}\right) &= \ln \left( e^{\enclose{circle}{\:k\enclose{circle}{\:t\:}\:}\:} \right) \\ \ln\left(\frac{P}{P_0}\right) &= \enclose{circle}{\:k\enclose{circle}{\:t\:}\:} \\ \end{align}

Finally, we can divide by $$k$$ and we have finished changing the subject to $$t$$.

\begin{align} \ln\left(\frac{P}{P_0}\right) &= \enclose{circle}{\:k\enclose{circle}{\:t\:}\:} \\ (\div k) &\:\:\:\:\:\ (\div k) \\ \frac{\ln\left(\frac{P}{P_0}\right)}{k} & = t \\ t & =\frac{\ln\left(\frac{P}{P_0}\right)}{k} \\ \end{align}

## Example 3

One thing to note is that closest is not in terms of distance but defined by the order of mathematical operations. Typically this order is denoted by the acronym BODMAS or BIDMAS. The equation below might serve as an example.

$y = e^{(x - 3)}$

Suppose we want to change the subject to $$x$$, then we start by circling it.

$y = e^{(\enclose{circle}{x} - 3)}$

Now we move to the closest term. Distance wise this is the $$e$$ but actually we need to sort out the brackets (B in BODMAS) first and so we circle the $$-3$$ first and then the $$e$$.

\begin{align} y &= e^{(\enclose{circle}{x} - 3)}\\ y &= e^{(\enclose{circle}{\enclose{circle}{x} - 3\:})}\\ y &= \enclose{circle}{\:\:e^{(\enclose{circle}{\enclose{circle}{x} - 3\:})}\:}\\ \end{align}

To change the subject we start (as above) by taking $$\ln$$ of both sides.

\begin{align} y &= \enclose{circle}{\:\:e^{(\enclose{circle}{\enclose{circle}{x} - 3\:})}\:}\\ \ln(y) &= \ln\left(e^{(\enclose{circle}{\enclose{circle}{x} - 3\:})}\right)\\ \ln(y) &= \left(\enclose{circle}{\enclose{circle}{x} - 3\:}\right) \\ (+3) \:\: & \:\:\:\:\:\:\:\:\:\: (+3) \\ \ln(y) + 3 &= x \end{align}

Finally we would normally read from left to right giving us

$x = \ln(y) + 3.$

Hopefully these examples help introduce an alternative way of setting out a changing the subject of the formula type problem.