If Pex = Qe-x  for all real values of x, which one of the following statements is true?

The correct option is P=Q=0.

To understand why this is correct, we analyze the given mathematical statement. We are given the following equation:

$P{e}^x=Q{e}^{-x}$

This equation is stated to be true for all real values of x. This means it is an identity, and we can choose any real number for $x$ to investigate the relationship between the constants $P$ and $Q$.

Step 1: Substitute a convenient value for x
Let us substitute $x=0$ into the equation:
$P{e}^0=Q{e}^{-0}$
Since any non-zero number raised to the power of 0 is 1 ($e^0=1$), this simplifies to:
$P\left(1\right)=Q\left(1\right)$
Which gives us our first relation:
$P=Q$

Step 2: Substitute another value for x
Now let us substitute another real value, for instance, $x=1$:
$P{e}^1=Q{e}^{-1}$
Using the relation $P=Q$ from Step 1, we can substitute $P$ in place of $Q$:
$Pe=P{e}^{-1}$
Now, we rearrange the equation by subtracting $P{e}^{-1}$ from both sides:
$Pe-P{e}^{-1}=0$
Factoring out $P$:
$P(e-e^{-1})=0$

Step 3: Solve for P and Q
For the product $P(e-e^{-1})=0$ to be equal to zero, at least one of the factors must be zero.
Since $e$ is Euler's number (approximately 2.718), we know that:
$e-e^{-1}\approx 2.718-0.368\ne 0$
Because the term $(e-e^{-1})$ is non-zero, we must have:
$P=0$
Substituting $P=0$ back into the relation $P=Q$ gives:
$Q=0$

Therefore, the only values for which the equality holds for all real values of $x$ are indeed $P=0$ and $Q=0$, which is represented as P=Q=0.