Solving the Exponential Equation $Pe^x = Qe^{-x}$

We are given the equation $Pe^x = Qe^{-x}$, 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:

   $ Pe^x = Qe^{-x} $

2. Multiply both sides by $e^x$ to eliminate the $e^{-x}$ term:

   $ Pe^x \cdot e^x = Qe^{-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 $Pe^x = Qe^{-x}$ for all real values of $x$.