[Algebra]
Problem: Consider the following system of equations in which all logarithms have base 10:

$$
\begin{aligned}
(\log x)(\log y)-3 \log 5 y-\log 8 x & =a \\
(\log y)(\log z)-4 \log 5 y-\log 16 z & =b \\
(\log z)(\log x)-4 \log 8 x-3 \log 625 z & =c
\end{aligned}
$$
If $a=-4, b=4$, and $c=-18$, solve the system of equations.

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.

Suppose that $a=-4, b=4$ and $c=-18$.

Then the last version of the system is

$$
\begin{aligned}
p q & =2 \\
q r & =12 \\
p r & =6
\end{aligned}
$$

Multiplying the three equations together gives $p^{2} q^{2} r^{2}=2 \cdot 12 \cdot 6=144$.

Since $(p q r)^{2}=144$, then $p q r= \pm 12$.

Therefore, $r=\frac{p q r}{p q}=\frac{ \pm 12}{2}= \pm 6$ and $p=\frac{p q r}{q r}=\frac{ \pm 12}{12}= \pm 1$ and $q=\frac{p q r}{p r}=\frac{ \pm 12}{6}= \pm 2$.

Therefore, the solutions to the last version of the system are $(p, q, r)=(1,2,6)$ and $(p, q, r)=(-1,-2,-6)$.

Converting back to the original variables, we see that the solutions to the original system when $(a, b, c)=(-4,4,-18)$ are $(x, y, z)=\left(10^{4}, 10^{3}, 10^{10}\right)$ and $(x, y, z)=\left(10^{2}, 10^{-1}, 10^{-2}\right)$.



Final answer: $(10^{4}, 10^{3}, 10^{10}),(10^{2}, 10^{-1}, 10^{-2})$