Having finished the work on Casey's computer, you attempt to save the changes and find that this is function is protected by a password. You click the "hint" option and are given the following: What exact value of $x$ satisfies $e^{2 x}-5 e^{x}-14=0$ ? If there is no solution, enter NA. Enclose arguments of functions in parentheses and include a multiplication sign between terms. For example, $c * \log (h)$. What value do you enter? \[ x= \]

Step 1 ：Let's make a substitution to simplify the equation. Let \(y = e^{x}\), then the equation becomes \(y^2 - 5y - 14 = 0\).

Step 2 ：This is a quadratic equation, which we can solve by factoring: \(y^2 - 5y - 14 = 0\) becomes \((y - 7)(y + 2) = 0\).

Step 3 ：Setting each factor equal to zero gives the solutions \(y = 7\) and \(y = -2\).

Step 4 ：However, since \(y = e^{x}\) and the exponential function \(e^{x}\) is always positive for real \(x\), \(y\) cannot be negative. Therefore, \(y = -2\) is not a valid solution.

Step 5 ：So, we only have \(y = 7\), and substituting \(y = e^{x}\) back in gives \(e^{x} = 7\).

Step 6 ：To solve for \(x\), we take the natural logarithm of both sides: \(x = \ln(7)\).

Step 7 ：So, the exact value of \(x\) that satisfies the equation \(e^{2 x}-5 e^{x}-14=0\) is \(\boxed{x = \ln(7)}\).