Which of the following functions is a one-to-one and onto function from $\mathbb{R}$ to $\mathbb{R}$?

For a function from $\mathbb{R}$ to $\mathbb{R}$ to be one-to-one and onto, it must:

• Be one-to-one: different inputs give different outputs
• Be onto: every real number can be achieved as an output

Let me simplify each function first using properties of exponentials and logarithms.

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Option 1: $e^{\ln(|x|)}$

Using the property $e^{\ln(a)} = a$, this simplifies to $|x|$.

Problem: $|-3| = 3$ and $|3| = 3$, so different inputs ($-3$ and $3$) give the same output ($3$). This is not one-to-one.

Also, the domain excludes $x = 0$ since $\ln(0)$ is undefined.

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Option 2: $e^{\ln(x)}$

This simplifies to $x$ for $x > 0$.

Problem: The function is only defined for positive $x$ (since $\ln(x)$ requires $x > 0$). The domain is not all of $\mathbb{R}$, so this cannot be a function from $\mathbb{R}$ to $\mathbb{R}$.

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Option 3: $\ln(e^x)$

Using the property $\ln(e^a) = a$, this simplifies to $x$.

Note that $e^x$ is always positive for any real $x$, so $\ln(e^x)$ is defined for all real numbers.

This is the identity function $f(x) = x$:

• One-to-one: Each input maps to a unique output ✓
• Onto: For any real number $y$, we can find $x = y$ such that $f(x) = y$ ✓
• Domain is all of $\mathbb{R}$ ✓

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Option 4: $\ln(e^{|x|})$

This simplifies to $|x|$.

Problem: Same as Option 1 - different inputs like $-2$ and $2$ give the same output, so not one-to-one. Also, since $|x| \geq 0$, we can never get negative outputs, so not onto.

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The answer is Option 3 because $\ln(e^x) = x$ is the identity function, which maps every real number to itself, making it both one-to-one and onto.