A brief refresher of some algebra

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2022-12-26

A review of some handy mathematical tricks

Written by: Ned

In an effort to avoid forgetting what I learned during my math degree, but mostly because I think math is pretty cool and I haven’t done enough of it lately, I plan to start working through some problems most days—though how often they’ll get posted remains to be seen. I haven’t planned out exactly where I’ll go with them, but I plan to draw on the material that was covered while I was studying either to retain what I learned or to deepen my understanding of topics I found interesting but didn’t have time to examine further. I will endeavour to avoid jumping straight into advanced math, but I make no guarantees that I’ll cover all of the prerequisites for a topic before I start working on it. At the moment though, I plan to at least touch on the basics and go from there.

I thought I’d start with a fairly straightforward problem that uses a fair amount of algebraic manipulation to solve if the inspected solution isn’t evident. So, without further ado let’s take a look at, $e^{x} + 5e^{-x} = 6$. By inspection, since $1 + 5 = 6$, since $e^0 = 1$ it’s fairly obvious that $x = 0$ is a solution, and similarly $e^{x} = 5$ or $x = \ln{5}$ must also be a solution (since $e^{-x}$ would then be $\frac{1}{5}$ and $5 + 5\frac{1}{5} = 6$ too). However, many equations of this type have less obvious solutions and but still crop up frequently enough that solving it algebraically is a worthwhile exercise even without the benefit it provides in reviewing exponents, simple logarithms, factoring, and the quadratic equation.

$e^{x} + 5e^{-x} = 6$

Splitting the exponent on $e^{-x}$ yields:

$e^{x} + 5(e^{x})^{-1} = 6$

$e^{x} + \frac{5}{e^{x}} = 6$

Multiplying the equation by $e^{x}$ gives

$e^{x}e^{x} + e^{x}\frac{5}{e^{x}} = 6e^{x}$

$e^{2x} + 5 = 6e^{x}$

$e^{2x} -6e^{x} + 5 = 0$

Using variable substitution and letting $e^{x}$ be $u$ gives us the quadratic:

$u^{2} -6u + 5 = 0$

Which factors into:

$(u - 1)(u - 5) = 0$

or, had it contained something less simple to factor, solve using the quadratic formula:

$$u = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$

So then noting that $a = 1$, $b = -6$, $c = 5$:

$u = \frac{6 \pm \sqrt{6^{2} - 4(1)(5)}}{(2)(1)}$

$u = \frac{6 \pm \sqrt{36 - 20}}{2}$

$u = \frac{6 \pm \sqrt{16}}{2}$

$u = \frac{6 \pm 4}{2}$

$u = \frac{6 - 4}{2}$ or $u = \frac{6 + 4}{2}$

$u = \frac{2}{2} = 1$ or $u = \frac{10}{2} = 5$

Thus, as was determined by initial inspection $u = 1$ or $u = 5$.

However, since $u = e^{x}$, we must solve for $x$ using the natural logarithm.

In this case taking the $\ln$ of both sides of our assignment statement:

$\ln{e^{x}} = \ln{u}$

$x = \ln{u}$

Then using our solved values of $u$.

$x = \ln{5}$ or $\ln{1} = 0 = x$

So the solutions to the equation $e^{x} + 5e^{-x} = 6$ are $x = 0$ and $x = \ln{5}$.

It’s always important to verify that the solutions are correct, and in this case at least, substituting the values one at a time into the original equation yields the correct result.

For $x = 0$

$e^{0} + 5e^{-0}$

$= 1 + 5(1)$

$= 6$

For $x = \ln{5}$

$e^{\ln5} + 5e^{-\ln5}$

$= e^{\ln5} + 5e^{\ln{5^{-1}}}$

$= 5 + (5)(5^{-1})$

$= 5 + 1$

$= 6$

Some discussion around technique

This particular equation was fairly straightforward to solve by inspection. But had the coefficients been less nice, it could as easily have been quite difficult to see how to solve it. The algebraic solution should work much more generally than inspection, but it’s nice to run through it with pleasant numbers and the algebraic tricks to solve it provide a nice review of various manipulations that occur frequently in more difficult problems. In particular, isolating and factoring or combining common terms ($e^x$ and $e^{-x}$) is quite useful. In this example, it required some knowledge of the exponent laws—in particular that $(a^{y})^{x} = a^{yx}$ to factor out the $-1$ and isolate $e^{x}$. In calculus, change of variables or u-substitution is frequent in solving integrals where the chain rule needs to be undone. In more complicated equations using symbols rather than complicated terms can reduce the number of transcription or arithmetic errors.

As ever, the quadratic equation is extremely useful and when combined with variable subsitution can frequently be used to solve exponential equations that have a quadratic form. Finally, the use of logarithms as an inverse function to an exponential term allowed us to find the solutions and easily evaluate the expression to determine that the solution was correct.