Q. \[ e^x = 0 \]

Q: Can e^x=0 have a real solution?

No. For all real x, e^x>0. Since the exponential function never reaches zero, the equation e^x=0 has no real solution.

Q: Why is e^x always positive?

Because e^x is defined so that e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}, and this series is always >0 for real x. Hence e^x\neq 0.

Q: Does solving e^x=0 ever make sense with logarithms?

Not directly. Taking logs requires a positive input: \ln(0) is undefined. Since e^x>0, there is no x making the right side 0.

Q: Are there any limits where e^x approaches 0?

Yes. \lim_{x\to -\infty} e^x = 0. This is a limit, not a solution to e^x=0.

Q: What is the smallest value of e^x?

There is no minimum. e^x can be arbitrarily small but never equal to 0. So the infimum is 0, but it is not attained.

Answer

We solve the equation \(e^x=0\).

The exponential \(e^x\) is always positive for every real number \(x\), so it can never equal \(0\).

Final result: There are no real solutions.

\[
\text{No real } x \text{ satisfies } e^x=0.
\]

Detailed Explanation

We are asked to solve the equation \(e^x = 0\).

Step 1: Recall a key fact about the exponential function.

The exponential function \(e^x\) is always positive for every real number \(x\). In symbols:

\[
e^x > 0 \quad \text{for all real } x.
\]

Step 2: Compare both sides of the equation.

The left-hand side \(e^x\) is always greater than \(0\), while the right-hand side is \(0\). So it is impossible for \(e^x\) to equal \(0\) for any real number \(x\).

Step 3: State the conclusion.

Therefore, the equation \(e^x = 0\) has no real solution.

Final Answer: \(\varnothing\) (no real \(x\) satisfies \(e^x = 0\)).

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Algebra FAQ

Can \(e^x=0\) have a real solution?

No. For all real \(x\), \(e^x>0\). Since the exponential function never reaches zero, the equation \(e^x=0\) has no real solution.

What about complex solutions to \(e^x=0\)?

No solutions in complex numbers either. The exponential function \(e^z\) is never zero for any complex \(z\).

Why is \(e^x\) always positive?

Because \(e^x\) is defined so that \(e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}\), and this series is always \(>0\) for real \(x\). Hence \(e^x\neq 0\).

Does solving \(e^x=0\) ever make sense with logarithms?

Not directly. Taking logs requires a positive input: \(\ln(0)\) is undefined. Since \(e^x>0\), there is no \(x\) making the right side \(0\).

Are there any limits where \(e^x\) approaches \(0\)?

Yes. \(\lim_{x\to -\infty} e^x = 0\). This is a limit, not a solution to \(e^x=0\).

What is the smallest value of \(e^x\)?

There is no minimum. \(e^x\) can be arbitrarily small but never equal to \(0\). So the infimum is \(0\), but it is not attained.

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