Question

Show that the logarithmic function $f:R\stackrel{+}{0}\to R\mathrm{given}\mathrm{by}f\left(x\right)={\log }_{a}x,a>0$ is a bijection.

Answer 1

f:R⁺→R given by $f\left(x\right)={\log }_{a}x$, a>0
Injectivity:
Let x and y be any two elements in the domain (N), such that f(x) = f(y).
f(x) = f(y)
$$\begin{gathered} {\log }_{a}x={\log }_{a}y \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R), such that f(x) = y for some element x in R⁺ (domain).
f(x) = y
$$\begin{gathered} {\log }_{a}x=y \\ \Rightarrow x=a^y\in R^{+} \end{gathered}$$
So, for every element in the co-domain, there exists some pre-image in the domain.
$\Rightarrow$f is onto.
Since f is one-one and onto, it is a bijection.