Logarithm simultaneous equation 
Let $(x_1, y_1,z_1)$ and $(x_2, y_2, z_2)$ - where $x_1\ge y_1\ge z_1$ and $x_2\le y_2\le z_2$  - be two triplets satisfying the following simultaneous equations:
$$
\begin{align}
\log_{10}(2xy)&=\log_{10}x\cdot\log_{10}y\\
\log_{10}(xy)&=\log_{10}z\cdot\log_{10}y\\
\log_{10}(2xz)&=\log_{10}x\cdot\log_{10}z
\end{align}
$$
Find $(x_1+y_1+z_1)^{x_2y_2z_2}$. (answer is an integer)

My attempt:
Put $a=\log_{10}(x),b=\log_{10}(y),c=\log_{10}(z),k=\log2$. We get:
$$
\begin{align}
k+a+b&=ab\tag{1}\\
a+b&=bc\tag{2}\\
k+a+c&=ac\tag{3}
\end{align}
$$
Subtracting (1) and (3), we get: $b-c=a(b-c)$. This implies $b=c$ or $a=1$. 
If $a=1$, we get $k=-1$ from (1) which is invalid. If $b=c=d$ (let), we get: $a=d^2-d$ (from (2)) and from (1), we have: $k=ad-d-a=(d^2-d)d-d-(d^2-d)=d^3-2d^2$.
But now we have a cubic in $d$ which I've no clue how to solve. What is the correct way to solve this problem?

Update: I'll be happy to offer any clarification on the question. Also, there may be a typo or two (my books are not 100% perfect), so if you're getting an integer answer by making a small adjustment, please feel free to share that. Thanks!

Too long for a comment.
I think that you did a good job and that the solution $b=c=\color{red}{x}$ is the one to consider (I changed notation on purpose).
You ended with the cubic equation $x^3-2x^2-k=0$ which you do not know how to solve.
Let us follow the steps given here for $a=1$, $b=-2$, $c=0$, $d=-k=$. So, we have $$\Delta=-k\,(27 k+32) <0$$ "then the equation has one real root and two non-real complex conjugate roots".
We also have $p=-\frac{4}{3}$ and $q=-k-\frac{16}{27} <0$
Using the hyperbolic method for one real root, we have
$$t_0=\frac{4}{3} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{27
   }{16}k\right)\right)$$ and $$x=t-\frac b {3a}=\frac 23+\frac{4}{3} \cosh \left(\frac{1}{3} \cosh ^{-1}\left(1+\frac{27
   }{16}k\right)\right)$$
So, back to your notations we have $b=c=x$ and $a=x^2-x$ 
Now, $?\cdots ?\cdots ?\cdots ?$ 
