Can I prove there is no real solution except $x=0, x=1$, without using the function $W(x)$?

Can I prove there is no real solution except $$x=0, x=1$$, without using the function $$W(x)$$?

And is it possible to do it without using calculus?

$$2^x=x+1.$$

Here is my attempts:

$$2^x>0 \Rightarrow x+1>0 \Rightarrow x>-1$$.

Now, I need to look at these intervals.

$$x\in (-1,0]; [0,1]; [1, \infty)$$

Maybe, it is easy to prove there is no real solution for $$x>1$$. Because , for $$x\to\infty$$, we get $$2^x>>x$$.

Problematic point is , $$x\in [0,1]$$ or $$x\in [-1,0]$$.

For $$0, we get $$2>2^x>1$$, which is correct ,because $$2>x+1>1$$ also true. So, this method doesn't work. I need more rigorous method.

Just another way $$:\quad$$ You have already ruled out $$x \leqslant -1$$. For others,

Case $$x\in (0, 1)$$ using Bernoulli's inequality, $$(1+1)^x < 1+x$$, so there is no solution

Case $$x \in (-1, 0)\cup (1, \infty)$$, again with Bernoulli's inequality we have $$(1+1)^x > 1+x$$.

Thus the only remaining points to check are $$x\in \{0, 1\}$$.

To prove that no solution exists in $$(0,1),$$ you may use the series representation for the exponential function (which can be derived without using calculus). If $$x\in (0,1)$$, we have \begin{align} 2^x &=1+\frac{\ln(2)}{1!}x+\frac{\ln^2(2)}{2!}x^2+...\\ &=1+x\bigg(\frac{\ln(2)}{1!}+\frac{\ln^2(2)}{2!}x+...\bigg)\\ &<1+x\bigg(\frac{\ln(2)}{1!}+\frac{\ln^2(2)}{2!}+...\bigg)\\ &=1+x\\ \end{align} Which shows that $$2^x\lt 1+x$$ for $$x\in (0,1)$$.

With calculus:

Lemma: If a function $$f$$ is $$n$$ times differentiable on an interval $$I$$, and $$f$$ has $$n+1$$ distinct zeros on $$I$$, then the $$n$$th derivative of $$f$$ has a zero somewhere in that interval.
Proof sketch: Apply Rolle's theorem repeatedly. There's a zero of $$f'$$ between each pair of zeros of $$f$$, for a total of at least $$n$$, then a zero of $$f''$$ between each zero of $$f'$$, and so on.

Now, for this problem? Consider $$f(x)=2^x-x-1$$. The second derivative is $$f''(x)=\ln^2(2)\cdot 2^x$$, which is positive on all of $$\mathbb{R}$$. With no zeros of the second derivative, the contrapositive of the lemma gives us that there are no more than two zeros of $$f$$ - on all of $$\mathbb{R}$$. We already know of two ($$0$$ and $$1$$), so there are no others.

• In the second sentence. The first sentence, and the title, was to do it without invoking the Lambert function. This post asked multiple questions, and I answered one of them. – jmerry Dec 7 '18 at 22:35

Function $$x\mapsto 2^x$$ is strictly convex.

A strictly convex intersects a line in 0, 1 or 2 points.

So there exists at most two solutions of the equation $$2^x = x +1$$.