Learning Lab

Examples

1. Make \(m\) the subject in the formula:11 In the formula, \(E\) is the subject because we have \(E=\cdots\). We want to rearrange to get \(m=\cdots\). \[ E=mgh+\frac{1}{2}mv^{2}. \] In this case the \(m\) ’s are already on the same side of the equation so we can take out \(m\) as a factor: \[ E=m\left(gh+\frac{1}{2}v^{2}\right). \] To get \(m\) on its own we divide both sides by \(gh+\frac{1}{2}v^{2}\) to get: \[\begin{align*} \frac{E}{gh+\frac{1}{2}v^{2}} & =\frac{m\left(gh+\frac{1}{2}v^{2}\right)}{gh+\frac{1}{2}v^{2}}\\ & =m\quad\left(\textrm{after cancelling}\right). \end{align*}\] That is, \[ m=\frac{E}{gh+\frac{1}{2}v^{2}}. \] It is possible to improve the look of the answer by getting rid of the \(1/2\) in the denominator. We do this by multiplying the top and bottom of the right hand side by \(2\) to get: \[\begin{align*} m & =\frac{E}{gh+\frac{1}{2}v^{2}}\times\frac{2}{2}\\ & =\frac{2E}{2gh+v^{2}}. \end{align*}\]

2. Make \(I\) the subject of the formula \(Ir=E-IR\).
   First get all terms involving \(I\) to one side by adding \(IR\) to both sides: \[\begin{align*} Ir+IR & =E-IR+IR\\ & =E. \end{align*}\] Take \(I\) out as common factor: \[\begin{align*} I\left(r+R\right) & =E.\\ \end{align*}\] Now divide both sides by \(r+R\): \[ I=\frac{E}{r+R}. \]

The examples above showed all steps in detail. You don’t have to do that when answering questions. For example 1 above you would write: \[\begin{align*} E & =mgh+\frac{1}{2}mv^{2}\\ & =m\left(gh+\frac{1}{2}v^{2}\right)\\ m & =\frac{E}{gh+\frac{1}{2}v^{2}} \end{align*}\]

For example 2 above you would write: \[\begin{align*} Ir & =E-IR\\ Ir+IR & =E\\ I\left(r+R\right) & =E\\ I & =\frac{E}{r+R} \end{align*}\]

Examples

1. For example, make \(u\) the subject of the formula \(1/f=1/v-1/u\). A suitable common divisor of the denominators is \(fvu\).33 A common divisor is the product of the denominators. For example to simplify \(\frac{1}{4}+\frac{1}{8}\) you can use a lowest common denominator of \(8\) or a common denominator of \(8\times4=32\). Either will give the same result, though the lowest common denominator involves smaller numbers. To simplify algebraic fractions like \(\frac{1}{v}-\frac{1}{u}\) we must use a common denominator \(v\times u=vu\) as we don’t know the values of \(v\) and \(u\). Multiplying both sides of the formula by \(fvu\) gives \[\begin{align*} \frac{1}{f}fvu & =\frac{1}{v}fvu-\frac{1}{u}fvu\\ vu & =fu-fv\quad\left(\textrm{after canceling the terms}\right). \end{align*}\] Next step is to get all terms involving \(u\) on the same side: \[\begin{align*} vu-fu & =-fv\quad\left(\textrm{subtracting $fu$ from both sides}\right). \end{align*}\] Now we factorize the left hand side: \[\begin{align*} u\left(v-f\right) & =-fv. \end{align*}\] The final step is to divide both sides by \(v-f\) to get \[\begin{align*} u & =\frac{-fv}{v-f}\\ & =-\frac{fv}{v-f}. \end{align*}\] This is a correct answer but having a \(-\) sign in front is ugly. We get rid of it by multiplying the top and bottom of the right hand side by \(-1\): \[\begin{align*} u & =\frac{-fv\left(-1\right)}{\left(v-f\right)\left(-1\right)}\\ & =\frac{fv}{\left(f-v\right)}. \end{align*}\] The example above showed all steps. You don’t have to provide such detail. An acceptable solution would be: \[\begin{align*} \frac{1}{f} & =\frac{1}{v}-\frac{1}{u}\\ vu & =fu-fv\\ u\left(v-f\right) & =-fv\\ u & =\frac{-fv}{v-f}\\ & =\frac{fv}{f-v}. \end{align*}\]

2. Transpose \(m=\sqrt{\frac{d-s}{s\left(e-f\right)}}\) to make \(s\) the subject.
   The square root sign acts as a bracket and we want to remove it so we can get at the subject term \(s\).44 The term \(\frac{d-s}{s\left(e-f\right)}\) is considered as a single term. We remove it by squaring both sides: \[\begin{align*} m^{2} & =\left(\sqrt{\frac{d-s}{s\left(e-f\right)}}\right)^{2}\\ & =\frac{d-s}{s\left(e-f\right)}. \end{align*}\] Now multiply both sides by \(s\left(e-f\right)\) to get:55 Note that \(m^{2}=m^{2}/1\) and so the common denominator of both sides is \(1\times s\left(e-f\right)=s\left(e-f\right)\). By multiplying by \(s\left(e-f\right)\)we remove fractions. \[\begin{align*} s\left(e-f\right)m^{2}= & d-s. \end{align*}\] Now we gather all the \(s\) terms to one side. We add \(s\) to both sides: \[\begin{align*} s\left(e-f\right)m^{2}+s= & d. \end{align*}\] Take a common factor of \(s\) out of the right hand side: \[\begin{align*} s\left(\left(e-f\right)m^{2}+1\right) & =d. \end{align*}\] Now we divide by \(\left(\left(e-f\right)m^{2}+1\right)\) to get the result \[\begin{align*} s & =\frac{d}{\left(\left(e-f\right)m^{2}+1\right)}. \end{align*}\] The example above showed all steps. You don’t have to provide such detail. An acceptable solution would be \[\begin{align*} m & =\sqrt{\frac{d-s}{s\left(e-f\right)}}\\ m^{2} & =\frac{d-s}{s\left(e-f\right)}\\ s\left(e-f\right)m^{2} & =d-s\\ s\left(e-f\right)m^{2}+s & =d\\ s\left(\left(e-f\right)m^{2}+1\right) & =d\\ s & =\frac{d}{\left(\left(e-f\right)m^{2}+1\right)}. \end{align*}\]

Examples

1. Make \(t\) the subject in the formula \(Q=Ae^{kt}.\)
   Take \(\log_{e}\)of both sides to get \[\begin{align*} \log_{e}Q & =\log_{e}\left(Ae^{kt}\right) \end{align*}\] Use log laws66 \(\log\left(ab\right)=\log_{e}a+\log_{e}b\). to simplify the right hand side: \[\begin{align*} \log_{e}Q & =\log_{e}A+\log_{e}\left(e^{kt}\right) \end{align*}\] Use log laws77 \(\log x^{n}=n\log x\). to simplify the second term on the right hand side:88 Remember \(\log_{e}e=1.\) \[\begin{align*} \log_{e}Q & =\log_{e}A+kt\log_{e}e\\ & =\log_{e}A+kt \end{align*}\] Now rearrange for \(kt\) and use log laws:99 \(\log b-\log a=\log\left(\frac{b}{a}\right).\) \[\begin{align*} kt & =\log_{e}Q-\log_{e}A\\ & =\log_{e}\left(\frac{Q}{A}\right). \end{align*}\] Finally divide both sides by \(k\) to get1010 Remember dividing by k is the same as multiplying by \(1/k\). \[ t=\frac{1}{k}\log_{e}\left(\frac{Q}{A}\right). \] This example shows all steps. You don’t have to provide such detail. An acceptable solution would be \[\begin{align*} \log_{e}Q & =\log_{e}\left(Ae^{kt}\right)\\ & =\log_{e}A+kt\\ kt & =\log_{e}Q-\log_{e}A\\ & =\log_{e}\frac{Q}{A}\\ t & =\frac{1}{k}\log_{e}\frac{Q}{A} \end{align*}\]

2. Make \(n\) the subject of the formula \(S=P\left(1+i\right)^{n}.\) In this example, we show explanatory notes on each line. You don’t have to provide these notes in your work.
   We have: \[\begin{align*} S & =P\left(1+i\right)^{n}\\ \log S & =\log\left[P\left(1+i\right)^{n}\right]\quad\left(\textrm{taking logs of both sides}\right)\\ & =\log P+\log\left(1+i\right)^{n}\quad\left(\textrm{using log laws}\right)\\ & =\log P+n\log\left(1+i\right)\quad\left(\textrm{using log laws}\right)\\ n\log\left(1+i\right) & =\log S-\log P\quad\left(\textrm{rearranging}\right)\\ & =\log\left(\frac{S}{P}\right)\quad\left(\textrm{using log laws}\right)\\ n & =\frac{\log\left(\frac{S}{P}\right)}{\log\left(1+i\right)}. \end{align*}\]

3. Transform the formula \(T=\frac{1}{c}\log_{e}\left(m-A\right)\) to make \(m\) the subject.1111 In this example we use the fact that \(cT=\log_{e}\left(m-A\right)\) is equivalent to \(e^{cT}=m-A.\) In this example, we show explanatory notes on each line. You don’t have to provide these notes in your work.
   We have: \[\begin{align*} T & =\frac{1}{c}\log_{e}\left(m-A\right)\\ cT & =\log_{e}\left(m-A\right)\quad\textrm{multiplying both sides by $c$ }\\ e^{cT} & =e^{\log_{e}\left(m-A\right)}\quad\textrm{exponentiating both sides using base $e$ }\\ e^{cT} & =m-A\quad\textrm{using log laws}\\ e^{cT}+A & =m\quad\textrm{adding $A$ to both sides}\\ m & =A+e^{cT}\quad\textrm{after rearranging.} \end{align*}\]