From the equation $a-b=5$, solve for one variable in terms of the other (for example, $b$ in terms of $a$) and substitute into the equation $ab=12$.

After substitution, you should get a quadratic equation in $a$. Use the quadratic formula and then think about which solution is possible for $e^x$, remembering that exponentials are always positive.

In Desmos, treat $a$ as the horizontal axis variable $x$ and $b$ as the vertical axis variable $y$. Enter the equations y = 12/x and y = x - 5 to represent $ab=12$ and $a-b=5$.

Tap the intersection points of the two graphs. There will be one intersection in the first quadrant (both coordinates positive) and one in the third quadrant (both negative). The $x$-coordinate of the first-quadrant intersection is the value of $e^x$.

Type each answer choice expression (for example, (5+sqrt(73))/2, (5-sqrt(73))/2, etc.) into new lines in Desmos. Compare their numerical values to the $x$-coordinate of the positive intersection; the one that matches is the correct choice for $e^x$.

We are given

Let $a=e^x$ and $b=e^y$. Then $a>0$ and $b>0$ because exponential values are always positive.

In terms of $a$ and $b$:

• $e^{x+y}=e^x\cdot e^y$ becomes $ab=12$.
• $e^x-e^y=5$ becomes $a-b=5$.

So the system is now

From $a-b=5$, solve for $b$ in terms of $a$:

Substitute this into $ab=12$:

Now we have an equation in one variable $a$.

Expand and rearrange $a(a-5)=12$:

Use the quadratic formula for $a^2-5a-12=0$, where $A=1$, $B=-5$, and $C=-12$:

So the two algebraic solutions for $a$ are $a = \dfrac{5+\sqrt{73}}{2}$ and $a = \dfrac{5-\sqrt{73}}{2}$.

Recall that $a=e^x$, and $e^x$ must be positive.

Compute the sign of each root:

• $\sqrt{73}$ is a little more than $8$, so $5+\sqrt{73} > 0$ and $\dfrac{5+\sqrt{73}}{2} > 0$.
• $5-\sqrt{73}$ is negative, so $\dfrac{5-\sqrt{73}}{2} < 0$.

Since $e^x>0$, we must take

Comparing with the choices, this matches option A.