Why does $\lambda (e^x - 1) = x$ have two solutions for $\lambda > 1$?

Question

Apparently $\lambda (e^x - 1) = x$ has two solutions for $\lambda > 1$.

My textbook is kind of handwaving and saying that this is true without explaining. Can somebody prove this for me or show me why this is true? I understand why $x=0$ is always a solution, but why will a new solution appear for $\lambda > 1$? I see that it is true when I graph it, but I'd like a more algebraic proof for it if possible.

Answer 1

Note that $$ \lambda(e^x - 1) = x \iff\\ e^x - \frac 1\lambda x - 1 = 0 $$ It's unfortunately impossible to "solve" for the roots of this equation using algebra. However, a little calculus goes a long way. Define $$ f(x) = e^x - \frac 1\lambda x - 1 = 0 $$ We can find the critical points of this function by setting the derivative equal to $0$: $$ f'(x) = e^x - \frac 1\lambda = 0 \implies\\x = -\log(\lambda) $$ After verifying that the critical point at $x = -\log(\lambda)$ is necessarily a global minimum and that $\lim_{x \to \pm \infty} f(x) = \infty$, we note that $$ f(-\log(\lambda)) = \frac 1\lambda - \frac{\log \lambda}{\lambda} - 1 = \frac{1-\lambda - \log(\lambda)}{\lambda} $$ Now, if $\lambda > 1$, then this minimum of $f$ is negative. This implies that as $x$ increases, the graph for $f(x)$ must "start" positive, cross the $x$-axis to reach its minimum, and then increase towards $+\infty$, crossing the $x$-axis once more.

That is, $f(x) = 0$ must have two solutions.

Answer 2

As you noticed, function $$f(x)=\lambda (e^x - 1) - x$$ is equal to $0$ if $x=0$ for any value of $\lambda$. As, said in other answers, the derivative $$f'(x)=\lambda e^x-1$$ cancels at $x=\log \left(\frac{1}{\lambda }\right)$ and at this point $$f(x)=-\lambda -\log \left(\frac{1}{\lambda }\right)+1$$ which is negative for $1<\lambda$ and the second derivative $$f''(x)=\lambda e^x$$ is positive, then it is a minimum if $\lambda >1$. So, since $e^x$ grows much faster than $x$, the function will have another intersection with the $x$ axis.

The second solution has an analytical expression in terms of Lambert function $$x=-\lambda -W\left(-e^{-\lambda } \lambda \right)$$ and its value will be more and more negative when $\lambda$ increases.

For sure, there is a simpler manner of looking at the solution of $f(x)=0$ since it corresponds to the intersection of the two curves $y=e^x$ and $y=1+\frac{x}{\lambda}$