We have $G=\left( 0,\infty \right)-\left\{ 1 \right\}$ and $x\circ y={{x}^{\ln y}}$. How do you demonstrate that $\forall x,y\in G$ then $x\circ y\in G$?

Hint: In this question we have been given with the domain $G=\left( 0,\infty \right)-\left\{ 1 \right\}$ and all the ordered pair relation of $x$ and $y$ have the value given as $x\circ y={{x}^{\ln y}}$. We have to prove that for all the values of $x$ and $y$ in the given domain of $G$, then the ordered pair relation of $x$ and $y$ belongs to the given domain $G$.

Complete step by step solution:
We have the given domain as $G=\left( 0,\infty \right)-\left\{ 1 \right\}$.
This represents the domain of $\left( 0,\infty \right)$ in which the number $1$ does not belong. This indicates that the domain has all positive numbers from $0$ to $\infty$ except the number $1$.
This means that for all $x$ belonging to the domain $G$, the ordered pair value $x\circ y={{x}^{\ln y}}>0$
Since the domain consists of only positive numbers.
Now since the domain does not consist the number $1$, we get $x\circ y\ne 1$.
Now suppose the contrary, let $x\circ y=1$
We can write it as:
$\Rightarrow {{x}^{\ln y}}=1$
On taking log on both sides, we get:
$\Rightarrow \ln \left( {{x}^{\ln y}} \right)=\ln \left( 1 \right)$
Now we know the property of logarithm that $\log {{a}^{b}}=b\log a$ therefore, we get the left-hand side of the expression as:
$\Rightarrow \ln y\ln x=\ln \left( 1 \right)$
Now we know that $\ln \left( 1 \right)=0$ therefore, on substituting, we get:
$\Rightarrow \ln y\ln x=0$
Now we know the mathematical property that when $ab=0$ either $a=0$ or $b=0$. On applying this property, we get:
$\Rightarrow \ln y=0$ or $\ln x=0$
Now for the value to be $0$, the value of $x$ and $y$ has to be $1$ since, $\log 1=0$.
Now this contradicts the fact that $x,y\in \left( 0,\infty \right)-\left\{ 1 \right\}$.
Therefore, we have showed that $\forall x,y\in G$ then $x\circ y\in G$, which is the required solution.

Note: It is to be remembered that when specific elements are written in subtraction form in the domain then those elements don’t belong in the domain. The property of logarithms should be remembered to simplify the expressions of this kind. It is to be noted that the log we used in the question is the natural log which has the base $e$.