Range of roots of the equation $e^{x}-ax-b=0$

Question

The equation $e^{x}-ax-b=0$

(A) one real root if $a\le0$

(B) one real root if $b > 0, a\le0$

(C) two real roots if $a > 0, a log_ea ≥a-b$

(D) no real root if $a > 0, a log_ea <a-b$

Answer 1

Since the wording of the question doesn't match the rules as pointed out by Martin R, I will not give an answer on usual form.

The calculus below is more formal. This is probably not the expected way, just for information:

$$e^x=ax+b$$ $$(ax+b)e^{-x}=1 $$ $$-(x+\frac{b}{a})e^{-x}=-\frac{1}{a}$$ $$-(x+\frac{b}{a})e^{-(x+\frac{b}{a})}=-\frac{1}{a}e^{-\frac{b}{a}}$$ Let $\quad Y=-(x+\frac{b}{a})\quad$ and $\quad X=-\frac{1}{a}e^{-\frac{b}{a}}$ $$Ye^Y=X$$ Solving this equation for $Y$ involves the Lambert W function : $$Y=W(X)$$ http://mathworld.wolfram.com/LambertW-Function.html

Thus the solution of $(1)$ is : $$-(x+\frac{b}{a})=W\left(-\frac{1}{a}e^{-\frac{b}{a}} \right)$$ $$\boxed{x=-\frac{b}{a}-W\left(-\frac{1}{a}e^{-\frac{b}{a}} \right)}$$ Knowing the properties of the Lambert W(X) function which is a multivalued function :

One real value for positive argument : $\quad -\frac{1}{a}e^{-\frac{b}{a}}>0$

Two real values for argument in range $-e^{-1}$ to $0$ that is $\quad -e^{-1}<-\frac{1}{a}e^{-\frac{b}{a}}<0$

No real value for argument $<-e^{-1}$.