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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 $0<x<1$, 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.

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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 $x\in \{0, 1\}$.

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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 $x\in (0,1)$.

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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.

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    $\begingroup$ 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. $\endgroup$ – jmerry Dec 7 '18 at 22:35
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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 $2^x = x +1$.

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