Equivalent Equations


Much of solving equations involves writing equations in equivalent forms (i.e., with the same solutions) that get progressively simpler in some sense. Certainly, one way to create an equivalent equation from some given one is through substitution of one expression seen in the equation with an equivalent one.

This has been a strategy we have employed many times already -- probably without even thinking about it -- in previous sections. As we have often seen, when the new expression is simpler than the previous one, the equation often gets easier to solve.

Frequently, such replacements are motivated by some rule or relationship we have previously discovered or learned. As quick examples:

Consider the following example, where a substitution is used in the first couple of steps:


Solve:     $\log_4 (x^2 + 2x + 1) = 6$


$$\begin{array}{rcll} \log_4 (x^2 + 2x + 1) &=& 6 & \quad {\Tiny \textrm{now we substitute } (x+1)^2 \textrm{ for } (x^2 + 2x + 1) \textrm{, via factoring a binomial squared}}\\ \log_4 (x+1)^2 &=& 6 & \quad {\Tiny \textrm{we can again make a substitution, given the log rule: } \log_b p^n = n\log_b}\\ 2\log_4 (x+1) &=& 6 & \quad {\Tiny \textrm{from here, we can use "socks and shoes"...}}\\ \log_4 (x+1) &=& 3 & \\ x+1 &=& 4^3 & \quad {\Tiny \textrm{note, we see the input to the preceding logarithm is positive, which is good -- we need it in the domain!}}\\ x &=& 4^3 - 1 & \\ x &=& 63 & \end{array}$$

Generally, making such substitutions is relatively safe. As a simple example, suppose we are solving $2m + 3m = 15$. We of course know that $2m + 3m = 5m$, and so we confidently first replace $2m+3m$ with $5m$ (i.e., we collect like terms). This produces the equivalent equation $5m=15$, which we then turn into a solution by dividing both sides by $5$, discovering $m=5$.

However, carefully thinking about domain issues as we solve equations is paramount, as issues can pop up with intended substitutions (and other actions taken, as the next section demonstrates) that on the surface look like they will create an equivalent equation -- but actually don't.

Consider what happens in the next example where we see an opportunity to replace a given expression with an equivalent one that presumably makes solving something easier.


Solve:     $\log_9 y^2 - \log_9 y = \log_9 (2y+5)$


One might remember the following property of logarithms (recalling the earlier section covering the "triangle of power" and logarithms): $$\log_x a - \log_x b = \log_x \cfrac{a}{b}$$ Thus, shouldn't the following be true? $$\log_9 y^2 - \log_9 y = \log_9 y$$ But then, can we not replace the $(\log_9 y^2 - \log_9 y)$ in our equation with $(\log_9 y)$, to then conclude $$\begin{array}{rcll} \log_9 y &=& \log_9 (2y+5) & {\Tiny \textrm{following this with an application of $9^x$ on both sides to eliminate the logarithms}}\\ y &=& 2y + 5 & {\Tiny \textrm{so we can subtract $2y$ from both sides to reduce the number of variables seen to one}}\\ -y &=& 5 & {\Tiny \textrm{now we can use "socks and shoes" from here}}\\ y &=& -5 & \end{array}$$

Alas, there is a problem!

Replacing $\log_9 y^2 - \log_9 y$ in our equation with $\log_9 y$ assumes that all of these expressions are well-defined. However, in the case when $y=-5$, the term $\log_9 y$ is not defined. $y=-5$ is not in the domain of the logarithm function!

As such, $y=-5$ is an extraneous solution and should be discarded.

There are no solutions to this equation.

Sneaky, right?

In general, if functions in an equation being solved have domains other than $\mathbb{R}$, one should always check all solutions produced to make sure the expressions in the orginal equation all exist.

However, substitutions are not the only way to create an equivalent equation from a given one...


Reversible Actions

Let us now explore actions that we can apply to both sides of an equation to produce an equivalent equation -- and other actions that don't, but how even these can still be helpful.

Many students will recognize statements like "add this value to both sides" or "multiply both sides by this" or "square both sides" from their experiences solving equations in the past. Most of these actions can be interpreted as applying the same function to both sides of an equation -- which provided there are no domain issues in that application, always produce another equation that is true when the first one was.

This will probably feel more clear when expressed this way:

If $a=b$ and their common value is in the domain of $f$, then $f(a) = f(b)$.

This probably feels obvious, and sure enough -- it's true. To convince any that might be worried, note that if $a$ is in the domain of $f$, then surely $f(a)=f(a)$. Then, because $a=b$, we can substitute a $b$ for the second $a$ to produce the desired result.

Now, recognize that for any equation that we might wish to solve (say, involving just the variable $x$) we can always find functions $g(x)$ and $h(x)$ so that we can write it as $g(x) = h(x)$. As an example, if we are trying to solve $$x^3 + \log_3 x = \sqrt{x^2 - 5^x}$$ Then, $$g(x) = x^3 + \log_3 x \quad \quad \textrm{and} \quad \quad h(x) = \sqrt{x^2 - 5^x}$$ Given the earlier result we highlighted above, it must then also be the case that:

If $x_0$ solves $g(x) = h(x)$ and $g(x_0) = h(x_0)$ is in the domain of $f$, then $x_0$ also solves $f(g(x)) = f(h(x))$.

As an example, we know that $x_0 = 2$ solves $2x - 3 = 1$. Noting that $1$ is in the domain of all of the following functions: $x + 5$, $7x$, $x^2$, $\sqrt[3]{x}$, $\log_9 x$ and $8^x$ we can deduce that $x_0=2$ solves all following too: $$\begin{array}{rcl} (2x-3)+5 &=& 1+5\\ 7(2x-3) &=& 7(1)\\ (2x-3)^2 &=& 1^2\\ \sqrt[3]{2x-3} &=& \sqrt[3]{1}\\ \log_9 (2x-3) &=& \log_9 1\\ 8^{2x-3} &=& 8^1 \end{array}$$ Importantly however, not all of these equations are equivalent to $2x-3=1$!

The problem lies with the third equation, $(2x-3)^2 = 1^2$, which is not only solved by $x_0=2$, but also by $x_1=1$ (since $x_1=1$ is not a solution to $2x-3=1$, these two equations are not equivalent).

Why did this happen? Think about it for a minute. We said a minute ago that $x_0$ was a solution to the first equation, then it would be a solution to the second. What we didn't say was that a solution $x_1$ to the second equation must also be a solution to the first. Applying a function to both sides only preserves existing solutions, it doesn't mean you can't pick up new ones!

So why did this not happen with all the other equations (which are all equivalent to $2x-3=1$)? Look at the nature of all these other functions applied. In all of these other cases, the function applied was invertible (i.e., it has an inverse).

When the action taken on one equation to produce another involves the application of an invertible function to both sides, that invertibility ensures the action is reversible -- and that in turn ensures the newly produced equation is equivalent ot the original one.

Let's look at a specific example. Consider the fourth case above, when the function applied to both sides is $\sqrt[3]{x}$. We know every solution to $2x-3=1$ must be a solution to $\sqrt[3]{2x-3} = \sqrt[3]{2x-3}$.

However, the inverse of taking a cube root is cubing. So now consider all of the solutions to the second equation. By the same logic, these must also be solutions to $(\sqrt[3]{2x-3})^3 = (\sqrt[3]{1})^3$, or equivalently: $2x-3 = 1$

Thus, any solution to either of these equations is a solution to the other -- they must then be equivalent equations!

When we apply a function to both sides that doesn't have an inverse -- like when we squared both sides (recall that $f(x) = x^2$ is not invertible when considering a domain of all reals) -- we can't construct the argument that ensures solutions to the second equation are also solutions to the first. As such, we then run the risk of picking up extraneous solutions.

Let's see how this plays out when we are actually trying to solve an equation:


Solve:     $\sqrt{x^2-3x-6} = x-1$


"Socks and shoes" can't be immediately applied as we have more than one $x$.

So let's try to be clever. We've seen that sometimes factoring can reduce the number of terms involving a variable, but the polynomial under the radical doesn't factor cleanly (Eisenstein's criterion, $p=3$), and the polynomial on the right is linear and doesn't factor any further.

Ok, that didn't work. Let's try another strategy. Note, if the radical wasn't there on the left side -- that would open the door to combining some of the terms involving $x$ together (via adding or subtracting something to both sides). Maybe that might result in fewer terms involving $x$? That seems promising, but how do we eliminate that radical?

Squaring both sides would certainly do it -- but we must be aware that as squaring is not invertible, so we might end up introducing one or more extraneous solutions if we do this. We'll consequently need to check any solutions we get to the equation resulting from squaring both sides to make sure they are also solutions to the original equation!$\require{color}$

$$\begin{array}{rcll} \sqrt{x^2-3x-6} &=& x-1 & {\color{red}{\Tiny \textrm{square both sides to get rid of the radical - but recognize this requires we check our final answers}}}\\ x^2-3x-6 &=& (x-1)^2 & {\Tiny \textrm{we won't be able to do much to reduce the number of $x$ terms unless we expand the right side}}\\ x^2-3x-6 &=& x^2-2x+1 & {\Tiny \textrm{subtracting $x^2$ on both sides (an invertible action) eliminates two terms involving $x$}}\\ -3x-6 &=& -2x+1 & {\Tiny \textrm{adding $2x$ to both sides (an invertible action) eliminates another $x$ term}}\\ -x - 6 &=& 1 & {\Tiny \textrm{we are down to a single occurrence of $x$, and can use socks and shoes from here}}\\ -x &=& 1+6 & \\ x &=& -(1+6) & {\Tiny \textrm{note, we can simplify along the way, or wait until the end to do so}}\\ x &=& -7 & \end{array}$$ But remember, we are NOT done yet! This solution for $x$ might only solve the squared equation, and not the original equation (i.e., it might be an extraneous solution). The simplest way to check if this is the case is to plug it into the left and right sides of the original equation (or any equation known to be equivalent to the original equation) and see if we get the same value.

In this case, we note $$\sqrt{(7)^2 - 3(-7) - 6} = 8 \quad \quad \textrm{while} \quad \quad (-7)-1 = -8$$ As $8 \neq -8$, we conclude that $x=-7$ was indeed extraneous. With no other solutions to consider, we have determined this equation actually has no solution.

Here's another example of taking an action on both sides of an equation that is not invertible and the consequences of doing so.


Solve:     $\cfrac{3x-4}{x-2} = \cfrac{2}{x-2}$


Note that we have three terms involving $x$, so "socks and shoes" isn't going to help us right away.

That said, notice the common denominator of $x-2$. If this common denominator had not involved a variable -- for example, if our equation had been $\frac{3x-4}{7} = \frac{2}{7}$, the equation would been easy to solve, with our first "socks and shoes" step involving multiplying both sides by $7$. Given that, would multiplying both sides by $(x-2)$ help in this case?

It seems like it would, but we need to be careful! What if, when we are all done, it turns out that $x=2$? This would have made $x-2 = 0$. That would mean that when we multiplied both sides by $(x-2)$, we actually multiplied both sides by zero! This is definitely not an invertible action!

We actually have two routes we could go from here...

We can go ahead and multiply both sides by $(x-2)$, but since doing so might be an action that is not invertible, we'll want to check any solutions we ultimately find back in the original equation. If there's going to be a problem related to the above, it will manifest as a zero in the denominator of that equation.

Alternatively, knowing a zero in the denominator will be a problem (which is true only when $x=2$), we could simply acknowledge that by appending the statement $(x \neq 2)$ to our work from this step forward.

Either approach is fine. We will use the second one here: $$\require{color}\begin{array}{rclll} \cfrac{3x-4}{x-2} &=& \cfrac{2}{x-2} & & \color{red}{\Tiny \textrm{multiply both sides by $(x-2)$, but recognize this is not an invertible action when $x=2$}}\\ 3x - 4 &=& 2 \quad & (x \neq 2) & {\Tiny \textrm{we are immediately down to one term involving $x$ -- we can use "socks and shoes" from here!}}\\ 3x &=& 2 + 4 \quad & (x \neq 2) & \\ x &=& (2+4)/3 \quad & (x \neq 2) & \\ x &=& 2 \quad & (x \neq 2) & \end{array}$$ However, realizing that $x=2$ and $x \neq 2$ can't both be true, we conclude that our original equation again has no solution!

Lest one be left with the impression that multiplying both sides by $(x-2)$ and then worrying about if this might have been a multiplication by zero is the only way to solve this equation -- let us quickly "nip that idea in the bud"!

In mathematics there is often more than one way to accomplish just about anything of interest! Consider the following (very different) strategy that reaches the same conclusion:


Solve:     $\cfrac{3x-4}{x-2} = \cfrac{2}{x-2}$


For any solution $x = x_0$, the two rational expressions in the equation will of course be defined and equal to some real value. As we know adding or subtracting any real value to both sides of an equation is an invertible action and will thus create an equivalent equation, let us subtract the rational expression on the right from both sides to find: $$\frac{3x-4}{x-2} - \frac{2}{x-2} = 0$$ Of course, we know how to collapse differences of rational expressions into a single rational expression. We need common denominators -- fortunately we already have them. Thus, $$\frac{3x-6}{x-2} = 0$$ From here, we might be tempted to reduce the fraction by factoring and "cancelling" any common factors found in the numerator and denominator. Be careful though -- this can have domain implications! For the moment, let's just factor things: $$\frac{3(x-2)}{x-2} = 0$$ Now we can see for $x=2$ the left side will be undefined, meaning $x=2$ is not a solution. For any other $x \neq 2$, the common factor of $(x-2)$ can be "canceled". But this would leave the equation $3=0$, one which is never true, regardless of the value of $x$. Consequently, we have (again) found no solutions to the original equation.

Recall when we discussed applying a function to both sides of an equation at the beginning of this section, we were explicit that the left and right sides must be in the domain of the function involved. This turns out to be important too!

For example, consider what happens if our initial equation was instead $2x+1=x$. We know $\log_9 x$ is an invertible function. Does this automatically mean that $\log_9 (2x+1) = \log_9 x$?

Nope! Notice that the first equation has solution $x_0=-1$, but this is going to create a domain isssue when we apply the log function to both sides, as we would be trying to take a logarithm of a negative value in both cases, which will not result in a real value.

Below is another example (also worked two ways) where we must carefully navigate domain issues:


Solve:     $(3x-6) \cdot \log_8 (x+7) = 3x-6$


We don't have a single occurrence of $x$, so we can't use "socks and shoes" just yet. However, one might see a common factor of $3x-6$ on both sides and be tempted to "cancel it" from both sides. This would result in only a single $x$ remaining, and then "socks and shoes" could be used.

However, to be more precise in our language -- when we "cancel" $3x-6$ from both sides, what we are actually doing is dividing both sides by $3x-6$. In doing so, we must be careful we aren't accidentally dividing by zero!

In this particular case, where $(3x-6)$ is a factor of both sides, also note that anytime $3x-6 = 0$ (and the logarithm is defined), we have found a solution to the equation as both the left and right sides of our equation will be zero for such $x$.

If we insist on dividing both sides by $(3x-6)$, we can do so -- but we'll have to break things into two cases given what we observed above -- what can you say if $3x-6 \neq 0$ and what can you say when $3x-6 = 0$:

Case 1: If $3x-6 \neq 0$, there is no problem dividing both sides by $3x-6$, so we do exactly that. After the division, what remains is the first equation below -- which upon seeing we now have only one occurrence of $x$, allows us to use "socks and shoes" from there:

$$\begin{array}{rcl} \log_8 (x+7) &=& 1\\ x+7 &=& 8^1\\ x &=& 8^1 - 7\\ x &=& 1 \end{array}$$ Case 2: If $3x-6 = 0$, we can't divide both sides by this value. Let's find out which $x$ values make this true by solving $3x-6=0$, which is easily also done with "socks and shoes": $$\begin{array}{rcl} 3x-6 &=& 0 \\ 3x &=& 0 + 6 &\\ x &=& (0+6)/3 &\\ x &=& 2 & \end{array}$$

Note that in this second case, just because $3x-6 = 0$ when $x=2$ doesn't mean that $x=2$ is automatically a solution to our original equation of $(3x-6) \cdot \log_8 (x+7) = 3x-6$. However, checking $x=2$ to see if it is a solution to that original equation is easy -- just plug it into the left and right sides to see if they agree. (Really, the only thing that could wrong at this point is for something (like the logarithm) to be undefined, as barring that both the left and right sides of our original equation will be zero.)

The left side is $0 \cdot \log_8 (2+7) = 0 \cdot \log_8 9 = 0$, while the right side is immediately $0$. As the left and right sides agree in value, $x=2$ must be a second solution to our original equation.

As such, we have two solutions to our equation: $x = 1$ or $x=2$.

As before, there is another (very different) strategy that could be employed as well -- one that minimizes the domain considerations one must make:


Solve:     $(3x-6) \cdot \log_8 (x+7) = 3x-6$


This time, let us manufacture a zero on one side of the equation with the idea that if we can factor the other side, then one of the factors must be zero. This generally produces multiple smaller (but hopefully easier) equations that can be solved instead of the original (presumably harder) one.

To get the zero on one side, let us subtract $(3x-6)$ from both sides. This is an invertible action, so we should get an equivalent equation:

$$(3x-6) \cdot \log_8 (x+7) - (3x-6) = 0$$ Our eyes are again drawn to the multiple occurrences of $(3x-6)$, which we now recognize as a common factor of the expression on the left. As such, $$(3x-6)(\log_8 (x+7) - 1) = 0$$ Again, the product of two real values can only be zero when at least one of them is zero. As such, we can solve for each possibility separately. In both cases, we can use "socks and shoes": $$ \begin{array}{ccc} \begin{array}[t]{rcl} 3x - 6 &=& 0\\ 3x &=& 6\\ x &=& 2 \end{array} & \quad \quad & \begin{array}[t]{rcl} \log_8 (x+7) - 1 &=& 0\\ \log_8 (x+7) &=& 1\\ x+7 &=& 8^1\\ x &=& 8^1 - 7\\ x &=& 1 \end{array} \end{array}$$ As such, we again find two solutions: $x=2$ and $x=1$.