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**Hint:**This is a typical logarithmic property question. We will need to use the property \[x\log x = \log \left( {{x^x}} \right)\] to get the final answer. Also we will need some properties of ratio to get started.

**Complete step-by-step answer:**It is given that \[\log x:\log y:\log z = (y - z):(z - x):(x - y)\]

Then we can let that,

\[\dfrac{{\log x}}{{y - z}} = \dfrac{{\log y}}{{z - x}} = \dfrac{{\log z}}{{y - z}} = k\]

So from here we can get

\[\begin{array}{l}

\log x = k(y - z)...................................(i)\\

\log y = k(z - x)................................(ii)\\

\log z = k(x - y).................................(iii)

\end{array}\]

Now if we multiply equation (i) with x, equation (ii) with y and equation(iii) with z

We will get its as

\[\begin{array}{l}

x\log x = kx(y - z)\\

y\log y = ky(z - x)\\

z\log z = kz(x - y)

\end{array}\]

So now let us add all three of them and we will get it as

\[\begin{array}{l}

\therefore x\log x + y\log y + z\log z = kx(y - z) + ky(z - x) + kz(x - y)\\

\Rightarrow x\log x + y\log y + z\log z = k\left( {xy - xz + yz - xy + yz + xz - yz} \right)\\

\Rightarrow x\log x + y\log y + z\log z = k \times 0\\

\Rightarrow x\log x + y\log y + z\log z = 0

\end{array}\]

Now by using the property of log, \[x\log x = \log \left( {{x^x}} \right)\] we will get the whole thing as

\[ \Rightarrow \log \left( {{x^x}} \right) + \log \left( {{y^y}} \right) + \log \left( {{z^z}} \right) = 0\]

Now again we know that \[\log x + \log y + \log z = \log (xyz)\]

So by using this property of log we are getting

\[ \Rightarrow \log \left( {{x^x}.{y^y}.{z^z}} \right) = 0\]

Again if we take the log to the other side we will get it as

\[ \Rightarrow \left( {{x^x}.{y^y}.{z^z}} \right) = 1\]

**Which is the correct answer and which means that option A is the correct option here.**

**Note:**\[\log x = 0\] is \[x = 1\] because we know that logarithm in mathematics is treated as log base e so here \[{\log _e}x = 0\] is basically \[x = {e^0} = 1\] . In other words we can also say that the antilog of 0 is 1.

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