If $P{e}^x=Q{e}^{-x}$ for all real values of $x$, which one of the following statements is true?

**Solving the Exponential Equation $P{e}^x=Q{e}^{-x}$ **

We are given the equation $P{e}^x=Q{e}^{-x}$, which must hold true for all real values of $x$. Our goal is to find the true statement about the constants $P$ and $Q$.

Algebraic Manipulation

1. Start with the given equation: $P{e}^x=Q{e}^{-x}$
2. Multiply both sides by $e^x$ to eliminate the $e^{-x}$ term: $P{e}^x\cdot e^x=Q{e}^{-x}\cdot e^x$
3. Apply the exponent rule $e^a\cdot e^b=e^{a+b}$: $P{e}^{x+x}=Q{e}^{-x+x}$ $P{e}^{2x}=Q{e}^0$
4. Simplify using $e^0=1$: $P{e}^{2x}=Q$

Determining Constants P and Q

The equation $P{e}^{2x}=Q$ must be true for every real number $x$.

- The term $e^{2x}$ varies as $x$ varies. For example, $e^{2(0)}=1$, $e^{2(1)}=e^2$, $e^{2(-1)}=e^{-2}$.
- Since $Q$ is a constant, it cannot change as $x$ changes.
- If $P$ were non-zero, the left side ($P{e}^{2x}$) would change value as $x$ changes, making it impossible to equal the constant $Q$ for all values of $x$.
- Therefore, for the equality $P{e}^{2x}=Q$ to hold true for all real $x$, the coefficient $P$ must be zero.
- Substituting $P=0$ into the equation gives $0\cdot e^{2x}=Q$, which simplifies to $0=Q$.
- Thus, the only condition that satisfies the original equation for all real $x$ is $P=0$ and $Q=0$.

Conclusion

The statement $P=Q=0$ is the only one that holds true given the initial condition $P{e}^x=Q{e}^{-x}$ for all real values of $x$.