Try letting $u=e^x$. What does $f(x)$ become in terms of $u$? What kind of function in $u$ is that?

For the quadratic in $u$, think about how to find its minimum value: you can use the vertex formula $u=-\dfrac{b}{2a}$ or complete the square. Which value of $u$ gives the smallest output?

Once you know the value of $u$ that minimizes the quadratic, remember that $u=e^x$. How do you solve $e^x = \text{(that number)}$ for $x$?

In Desmos, type f(x) = e^(2x) - 6e^x + 5 (you can also use exp(2x) - 6exp(x) + 5) to graph the function.

Tap or click on the curve and use the “minimum” feature (or just move along the graph) to find the lowest point of the graph. Desmos will display the coordinates $(x,y)$ of this minimum; note the $x$-value.

In new lines in Desmos, type each option: ln(2), ln(1.5), ln(6), and ln(3). Compare their decimal values to the $x$-coordinate of the minimum; the choice whose value matches that $x$-coordinate is the correct answer.

Notice that $e^{2x}=(e^x)^2$, so the function

can be seen as a quadratic expression in $e^x$.

Let $u=e^x$. Then $u>0$ and

Now the problem is: for what value of $u$ does $u^2-6u+5$ attain its minimum?

To find the minimum of the quadratic $u^2-6u+5$, it is easiest to complete the square:

The term $(u-3)^2$ is always nonnegative and is smallest when $(u-3)^2=0$, which happens when $u=3$. So the quadratic reaches its minimum value when $u=3$.

Recall that we set $u=e^x$. We found that the minimum occurs at $u=3$, so

To solve for $x$, take the natural logarithm of both sides:

Therefore, $f(x)$ attains its minimum when $x=\ln 3$, which corresponds to answer choice D.