Given the function $f(x) = e^x - ax$ solve for $a$ such that $f(x) \geq 1$. So I have the function $f: \mathbb{R} \to \mathbb{R}$, $f(x) = e^x - ax$ and it is known that $a > 0$. I need to find $a$ such that $f(x) \geq 1$, $\forall x\in \mathbb{R}$. 
What I have done so far is to set the derivative equal to $0$ in hopes of a minimum. So I found:
$f'(x) = e^x - a$
Set it equal to $0$ and found that the point $x = ln(a)$ is the global minimum point. 
Logic led me to think that since I have to find an $a$ for which the function is $\geq 0$ for all values of $x$, all I have to do is find the $a$ for which the function is $\geq 0$ at that minimum point. Since the minimum point is $x = ln(a)$, I have to solve:
$f(ln(a)) \geq 1$
That gives me 
$a - a*ln(a) \geq 1$, or
$a(1 - ln(a)) \geq 1$. 
Here is where I got stuck. Is my reasoning correct? Should I have done something differently? Is there a better way to do this? And if what I have done so far is correct, how could I go about solving for $a$?

Your reasoning is correct. It may help to graph the function $x(1-\ln(x))$ to see where it is at least 1. You will see a very clear critical point which should give you the solution to the problem graphically. You can then prove that this point is the only value for $a$ giving $f(x) \geq 1$. 
Now note that $a(1-\ln(a))$ is our minimum value of $f$ for any given $a$. We can treat this as a function and take its derivative wrt $a$. We get $a(-1/a)+(1-\ln(a))=-\ln(a)$. Setting this equal to zero gives $a=1$. This is a maximum by the second derivative test. Therefore the largest our minimum value for $f$ could be is its value at $a(1-\ln(a))$ where $a=1$, which is $1$. Since this is a maximum, all other minimum values for $f$ are less than 1, while when $a=1$ the minimum value is 1.
Note this does not work for nonpositive $a$ since that is outside the domain of ln. Can you find another argument which works for nonpositive $a$? 
