Subfield: Algebra


Question: Consider the following system of equations in which all logarithms have base 10:

$$
\begin{aligned}
(\log x)(\log y)-3 \log 5 y-\log 8 x & =a \\
(\log y)(\log z)-4 \log 5 y-\log 16 z & =b \\
(\log z)(\log x)-4 \log 8 x-3 \log 625 z & =c
\end{aligned}
$$
Determine all triples $(a, b, c)$ of real numbers for which the system of equations has an infinite number of solutions $(x, y, z)$.

Solution:
Using $\log$ arithm rules $\log (u v)=\log u+\log v$ and $\log \left(s^{t}\right)=t \log s$ for all $u, v, s>0$, the first equation becomes

$$
\begin{aligned}
(\log x)(\log y)-3 \log 5-3 \log y-\log 8-\log x & =a \\
(\log x)(\log y)-\log x-3 \log y-\log 8-\log 5^{3} & =a \\
(\log x)(\log y)-\log x-3 \log y-\log (8 \cdot 125) & =a \\
(\log x)(\log y)-\log x-3 \log y-\log (1000) & =a \\
(\log x)(\log y)-\log x-3 \log y-3 & =a
\end{aligned}
$$

Similarly, the second equation becomes

$$
\begin{aligned}
(\log y)(\log z)-4 \log 5-4 \log y-\log 16-\log z & =b \\
(\log y)(\log z)-4 \log y-\log z-4 \log 5-\log 16 & =b \\
(\log y)(\log z)-4 \log y-\log z-\log \left(5^{4} \cdot 16\right) & =b \\
(\log y)(\log z)-4 \log y-\log z-\log (10000) & =b \\
(\log y)(\log z)-4 \log y-\log z-4 & =b
\end{aligned}
$$

And the third equation becomes

$$
\begin{aligned}
(\log z)(\log x)-4 \log 8-4 \log x-3 \log 625-3 \log z & =c \\
(\log z)(\log x)-4 \log x-3 \log z-4 \log 8-3 \log 625 & =c \\
(\log z)(\log x)-4 \log x-3 \log z-\log \left(8^{4} \cdot 625^{3}\right) & =c \\
(\log z)(\log x)-4 \log x-3 \log z-\log \left(2^{12} \cdot 5^{12}\right) & =c \\
(\log z)(\log x)-4 \log x-3 \log z-12 & =c
\end{aligned}
$$

Since each of the steps that we have made are reversible, the original system of equations is equivalent to the new system of equations

$$
\begin{aligned}
(\log x)(\log y)-\log x-3 \log y-3 & =a \\
(\log y)(\log z)-4 \log y-\log z-4 & =b \\
(\log z)(\log x)-4 \log x-3 \log z-12 & =c
\end{aligned}
$$

Next, we make the substitution $X=\log x, Y=\log y$ and $Z=\log z$. (This is equivalent to saying $x=10^{X}, y=10^{Y}$ and $z=10^{Z}$.)

This transforms the system of equations to the equivalent system

$$
\begin{aligned}
X Y-X-3 Y-3 & =a \\
Y Z-4 Y-Z-4 & =b \\
X Z-4 X-3 Z-12 & =c
\end{aligned}
$$

We re-write the first of these three equations as $X(Y-1)-3 Y-3=a$ and then as $X(Y-1)-3(Y-1)-6=a$ and then as $(X-3)(Y-1)=a+6$.

In a similar way, we re-write the second and third of these equations to obtain the equivalent system

$$
\begin{aligned}
(X-3)(Y-1) & =a+6 \\
(Y-1)(Z-4) & =b+8 \\
(X-3)(Z-4) & =c+24
\end{aligned}
$$



Next, we make the substitution $p=X-3, q=Y-1$ and $r=Z-4$. (This is equivalent to saying $X=p+3, Y=q+1$ and $Z=r+4$, or $x=10^{p+3}, y=10^{q+1}$ and $z=10^{r+4}$.)

This transforms the original system of equations into the equivalent system

$$
\begin{aligned}
p q & =a+6 \\
q r & =b+8 \\
p r & =c+24
\end{aligned}
$$

We again note that this system of equations is equivalent to the initial system of equations, and each solution of this system corresponds with a solution of the initial system.

We consider the various possibilities for the product, $(a+6)(b+8)(c+24)$, of the right sides of the equations in the final form of the system above: whether it is positive, negative or equal to 0 .

Case 1: $(a+6)(b+8)(c+24)<0$

As in (a), we multiply the three equations together to obtain $(p q r)^{2}=(a+6)(b+8)(c+24)$. Since the left side is at least 0 and the right side is negative, then there are no solutions to the system of equations in this case.

Case 2: $(a+6)(b+8)(c+24)>0$

As in (a), we multiply the three equations together to obtain $(p q r)^{2}=(a+6)(b+8)(c+24)$. Since $(p q r)^{2}=(a+6)(b+8)(c+24)$ and $(a+6)(b+8)(c+24)>0$, then $p q r= \pm \sqrt{(a+6)(b+8)(c+24)}$.

Since $(a+6)(b+8)(c+24)>0$, then $\sqrt{(a+6)(b+8)(c+24)}$ is well-defined.

Also, since $(a+6)(b+8)(c+24)>0$, then each of $a+6, b+8, c+24$ is non-zero, so we can divide by each of these quantities.

As we did in (a), we can solve to obtain

$$
\begin{aligned}
& p=\frac{p q r}{q r}=\frac{ \pm \sqrt{(a+6)(b+8)(c+24)}}{b+8} \\
& q=\frac{p q r}{p r}=\frac{ \pm \sqrt{(a+6)(b+8)(c+24)}}{c+24} \\
& r=\frac{p q r}{p q}=\frac{ \pm \sqrt{(a+6)(b+8)(c+24)}}{a+6}
\end{aligned}
$$



Since $(a+6)(b+8)(c+24)>0$, these are all valid fractions and there are exactly two triples $(p, q, r)$ that are solutions and so two triples $(x, y, z)$ that are solutions to the original system.

Case 3: $(a+6)(b+8)(c+24)=0$

$\text { Suppose that exactly one of } a+6, b+8$ and $c+24$ equals 0 .

Without loss of generality, suppose that $a+6=0, b+8 \neq 0$ and $c+24 \neq 0$.

Since $p q=a+6=0$, then $p=0$ or $q=0$.

In this case, either $q r=b+8$ or $p r=c+24$ will equal 0 , which contradicts our assumption that neither $b+8$ nor $c+24$ is 0 .

Therefore, it cannot be the case that exactly one of $a+6, b+8$ and $c+24$ equals 0 .

Suppose next that exactly two of $a+6, b+8$ and $c+24$ equal 0 .

Without loss of generality, suppose that $a+6=b+8=0$ and $c+24 \neq 0$.

Since $p r=c+24 \neq 0$, then $p \neq 0$ and $r \neq 0$.

Since $p q=a+6=0$ and $q r=b+8=0$ and $p \neq 0$ and $r \neq 0$, then $q=0$.

In this case, any triple $(p, q, r)$ with $q=0$ and $p r=c+24 \neq 0$ is a solution to the system of equations.

Thus, when $a+6=b+8=0$ and $c+24 \neq 0$ (that is, $(a, b, c)=(-6,-8, c)$ with $c \neq 24)$, each triple $(p, q, r)=\left(p, 0, \frac{c+24}{p}\right)$ with $p \neq 0$ is a solution to the system of equations.

Each of these solutions corresponds to a solution to the original system of equations in $(x, y, z)$, so if $(a, b, c)=(-6,-8, c)$ with $c \neq 0$, then there are infinite number of solutions to the system of equations.

Similarly, if $(a, b, c)=(-6, b,-24)$ with $b \neq-8$ (that is, if $p=a+6=0$ and $r=c+24=0$ but $q=b+8 \neq 0)$ or $(a, b, c)=(a,-8,-24)$ with $a \neq-6$, then there are infinitely many solutions $(x, y, z)$ to the original system of equations.

Finally, we must consider the case of $a+6=b+8=c+24=0$.

Here, we must solve the system of equations

$$
\begin{aligned}
& p q=0 \\
& q r=0 \\
& p r=0
\end{aligned}
$$

Each triple $(p, q, r)=(0,0, r)$ is a solution of this system and there are infinitely many such solutions. (This is not all of the solutions, but represents infinitely many solutions.) Therefore, when $(a, b, c)=(-6,-8,-24)$, there are also infinitely many solutions to the original system of equations.

Therefore, the system of equations has an infinite number of solutions $(x, y, z)$ precisely when $(a, b, c)=(-6,-8, c)$ for some real number $c$ or $(a, b, c)=(-6, b,-24)$ for some real number $b$ or $(a, b, c)=(a,-8,-24)$ for some real number $a$ or $(a, b, c)=(-6,-8,-24)$. (This last triple is in fact included in each of the previous three families of triples.)

Final answer: $(a, b, c)=(-6,-8, c)$ for some real number $c$ or $(a, b, c)=(-6, b,-24)$ for some real number $b$ or $(a, b, c)=(a,-8,-24)$ for some real number $a$