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Which of the following are injections, or surjections or bijective? Justify your answer. $f:\left( 0,\infty \right)\to \text{R}$ defined by $f\left( x \right)={{\log }_{e}}x$.

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Last updated date: 21st Jul 2024
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Answer
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Hint: We must first assume two variables ${{x}_{1}}\text{ and }{{x}_{2}}$ is the domain of given function, such that $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$. If we get ${{x}_{1}}={{x}_{2}}$, then the function is injective. Again, assuming a variable ${{y}_{1}}$ in codomain, such that $f\left( {{x}_{1}} \right)={{y}_{1}}$, and if we get that for every ${{y}_{1}}$, the corresponding ${{x}_{1}}$ lies in the domain, then the function is surjective. We must also know that if the function is both injective and surjective, it is called bijective function.

Complete step-by-step solution:
Let us assume two values ${{x}_{1}},{{x}_{2}}\in \left( 0,\infty \right)$ such that $f\left( {{x}_{1}} \right)=f\left( {{x}_{2}} \right)$.
Using the definition of this given function, we can write
${{\log }_{e}}{{x}_{1}}={{\log }_{e}}{{x}_{2}}$
We can also write this as
${{\log }_{e}}{{x}_{1}}-{{\log }_{e}}{{x}_{2}}=0$
Using the property of logarithms, we can write
${{\log }_{e}}\dfrac{{{x}_{1}}}{{{x}_{2}}}=0$
On taking exponential on both sides, we can write
${{e}^{{{\log }_{e}}\dfrac{{{x}_{1}}}{{{x}_{2}}}}}={{e}^{0}}$
Again, using the property of logarithms, we can write
$\dfrac{{{x}_{1}}}{{{x}_{2}}}=1$
And so, we get
${{x}_{1}}={{x}_{2}}$.
Hence, the given function is one-one. Or, we can say that the given function is injective.
Let us now assume a variable ${{y}_{1}}\in \text{R}$, such that $f\left( {{x}_{1}} \right)={{y}_{1}}$.
Thus, we get
${{\log }_{e}}{{x}_{1}}={{y}_{1}}$
Using the definition of logarithms, we can write
${{x}_{1}}={{e}^{{{y}_{1}}}}$
We can see here that for every ${{y}_{1}}\in \text{R}$, the corresponding ${{x}_{1}}\in \left( 0,\infty \right)$.
Thus, we can say that the function is onto or surjective.
We now know that the given function is both injective and surjective.
Hence, we can say that the function is bijective.

Note: We must carefully include all the boundary conditions while checking the function for injectivity and for surjectivity. We must also be very clear that the injective function is also known as one-one function, and the surjective function is also called an onto function.