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To prove that $f$ is injective, we need to show that if $f(x_1) = f(x_2)$, then $x_1 = x_2$ for all $x_1, x_2 \in R^{+}$.
Let $f(x_1) = f(x_2)$. Then, $\log_{a} x_1 = \log_{a} x_2$.
Since the logarithm function is one-to-one, we can say that if $\log_{a} x_1 = \log_{a} x_2$, then $x_1 = x_2$.
Therefore, $f$ is injective (one-to-one).
To prove that $f$ is surjective, we need to show that for every $y \in R$, there exists an $x \in R^{+}$ such that $f(x) = y$.
Let $y \in R$. We want to find an $x \in R^{+}$ such that $f(x) = y$, i.e., $\log_{a} x = y$.
We can rewrite the equation $\log_{a} x = y$ as $x = a^y$.
Since $a > 0$, $a^y > 0$ for all $y \in R$. Thus, $x = a^y \in R^{+}$.
Therefore, for every $y \in R$, there exists an $x = a^y \in R^{+}$ such that $f(x) = y$. Hence, $f$ is surjective (onto).
Since $f$ is both injective and surjective, $f$ is a bijection.
Final Answer: f is a bijection
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