What is the value of $\log_b m^{\log_b n}$? What is the value of the following expression?

$$\log_b \left( m^{\log_b n} \right)$$

As far as I know it should be:
$$\log_bn\;\times\;\log_bm$$
Can it be simplified further? If so how?
Also, I read somewhere that it is equal to:
$$\log_b m$$
If that is correct, how is it possible?
Also, if it matters, the place where I read about the second answer, used base $10$.
Here's the image of the book where I found it:

 A: I'm assuming you mean $\log_b\left( m^{\log_bn} \right).$
Then $\log_bn\;\times\;\log_bm$ is correct, and that cannot be simplified any further. Perhaps you are thinking about $\log_b(n \times m)=\log_bn+\log_bm$?
And it cannot equal $\log_bm$ - where would the $n$ have gone?

EDIT to explain recently introduced picture in question:
$\log u=\log\frac{z}{y} \log x+\log\frac{x}{z} \log y+\log\frac{y}{x} \log z=(\log z-\log y)\log x+(\log x-\log z)\log y+(\log y-\log x)\log z=\log z \log x-\log y\log x+\log x\log y-\log z\log y+\log y\log z-\log x\log z=0$
A: Wow, I don't like that book at all.
It is true that
$$
\log u  = \log x^{\log (z/y)} + \log y^{\log(x/z)} + \log z^{\log(y/x)} 
$$
implies
$$
\log u = \log \frac zy + \log \frac xz + \log \frac yx,
$$
because the right-hand formulas in both equations are equal to zero,
but since what we need to do is to to show that the right-hand side
of the first equation is zero, we can't legitimately use that fact to
argue this implication.
It certainly is not true that in general
$\log x^a + \log y^b + \log z^b = a + b + c,$
so I really can't see how that step of the solution was justified.
I think a better approach is never to write the expression
$x^{\log (z/y)} y^{\log(x/z)}  z^{\log(y/x)}$ at all.
Instead, go back one step in the chain of equations $(1)$ in the book's
"solution", where we see the equal expression
$x^{\log z - \log y} y^{\log x - \log z}  z^{\log y - \log x}$.
That's an easier expression to work with.
Set
\begin{align}
u &= x^{\log z - \log y} y^{\log x - \log z} z^{\log y - \log x}, \\
\log u &= \log \left(x^{\log z - \log y} y^{\log x - \log z}
                  z^{\log y - \log x}\right) \\
  &= \log \left(x^{\log z - \log y}\right)
       + \log\left(y^{\log x - \log z}\right)
       + \log\left(z^{\log y - \log x}\right) \\
 &= (\log z - \log y)\log x + (\log x - \log z)\log y
                  + (\log y - \log x)\log z \\
 &= \log z \log x - \log y \log x + \log x \log y - \log z \log y
                  + \log y \log z - \log x \log z \\
 &= 0.
\end{align}
