## Problem 1

Let the set $G=(k, \infty) \backslash\{k+1\}, k>0$, on which the operation * is defined such that:

$\log _{a}[(x * y)-k]=\log _{a}(x-k) \cdot \log _{a}(y-k), \forall x, y \in G, a>0, a \neq 1$.

a) Determine the number of pairs of integers $(k, a)$ for which the neutral element of the operation * is $e=2014$.

b) Verify if the operation * is associative and determine the element $x \in G$ that satisfies the relation $x=x^{\prime}$, in the case $k=a+1=1005$. (The symbol $a^{\prime}$ denotes the inverse of $a$ with respect to the operation *).
## Problem 1

Let the set $G=(k, \infty) \backslash\{k+1\}, k>0$, on which the law $*$ is defined such that: $\log _{a}[(x * y)-$ $k]=\log _{a}(x-k) \cdot \log _{a}(y-k), \forall x, y \in G, a>0, a \neq 1$.

a) Determine the number of pairs of integers $(k, a)$ for which the neutral element of the law $*$ is $e=2014$.

b) Verify if the law * is associative and determine the element $x \in G$ that satisfies the relation $x=x^{\prime}$, in the case $k=a+1=1005$ (denoted by $a^{\prime}$ the symmetric of $a$ with respect to the law $*$ ).

## Solution

From the calculation, we obtain: $x * y=k+(x-k)^{\log _{a}(y-k)}, x, y \in G$.

1 point

a) From $e \in G$ being the neutral element, we get $x * e=e * x=x,(\forall) x \in G ; x * e=x \Leftrightarrow k+$ $(x-k)^{\log _{a}(e-k)}=x$

$\Leftrightarrow(x-k)^{\log _{a}(e-k)}=x-k \Leftrightarrow \log _{a}(e-k)=1 \Leftrightarrow e=k+a \in G \quad$.....2 points

$k+a=2014, k, a \in Z, k>0 \Rightarrow(k, a) \in\{(1,2013),(2,2012),(3,2011), \ldots,(2013,1)\}$, in total 2013 pairs. 1 point

b) Associativity is verified for $k=1005, a=1004$. 1 point

$x * x^{\prime}=e \Leftrightarrow x^{\prime}=k+a^{\frac{\log _{a}(e-k)}{\log _{a}(x-k)}} \in G \ldots \ldots \ldots . . .1$ point

For $x=x^{\prime}$ we obtain $\log _{1004}^{2}(x-1005)=1 \Rightarrow \log _{1004}(x-1005)=1 \Rightarrow x_{1}=2009 \in G$ or $\log _{1004}(x-1005)=-1 \Rightarrow x_{2}=1005+\frac{1}{1004} \in G \ldots \ldots . . .1$ point