Solving the equation $e^{x}-2^{x}=1$ I'm trying to solve
$e^{x}-2^{x}=1$
This came from one of my HS students, who was hoping to solve it algebraically. We graphed the first term vs the 2nd and third and see -

Which helped to understand there is a solution, and only one. By simply substituting 1 and then 2 for X, we had already concluded, but graphed for confirmation. Aside from staring at this for the last few days, we are not finding even a first step beyond this. The student is a sophomore, but our curriculum goes to AP Calculus, so if the answer is that there is no algebraic solution, but we need calculus to solve, that's fine.
He got this from a friend, and offered no background beyond that.
 A: By inspection, it is clear that the solution is just above $x=1$. Using Taylor series
$$e^x-2^x=\sum_{n=0}^\infty \frac{e-2 \log ^n(2)}{n!} (x-1)^n$$
Truncate to some order and use series reversion
$$x=1+t+\frac{t^2 \left(2 \log ^2(2)-e\right)}{2 (e-2\log (2))}+O\left(t^3\right)\qquad \text{with} \qquad t=\frac{3-e}{e-2 \log (2)}$$
This very truncated series gives $x=1.18199$; adding one more term would give $x=1.18780$.
A bit better would be to look at the zero of function
$$f(x)=x-\log(1+2^x)$$
$$f(x)=(1-\log (3))+(x-1) \left(1-\frac{2 \log (2)}{3}\right)-\frac{1}{9} (x-1)^2 \log^2(2)+O\left((x-1)^3\right)$$ and series reversion gives
$$x=1+u-\frac{u^2 \log ^2(2)}{6\log (2)-9}+O\left(u^3\right)\qquad \text{with} \qquad u=\frac{3 (\log (3)-1)}{3-2 \log (2)}$$
This gives $x=1.18666$ while Newton method would give $x=1.18674$
A: As noticed we need to use numerical methods or Lambert W-function to obtain the solution but as a first step we should prove rigorously that only one solution exists (note that the graph can give an insight but it doesn't suffices as a proof).
At this aim let consider
$$f(x)=e^x-2^x-1 \implies f'(x) =e^x-2^x\log 2>0$$
indeed
$$e^x-2^x\log 2>0 \iff 2^{x\log_2 e}-\frac{2^x}{\log_2 e}\iff (\log_2 e)2^{x\log_2 e}-2^x>0$$
therefore since $f(1)=e-3<0$ and $f(2)=e^2-5>0$ by IVT only one root exists in the interval $x\in(0,2)$.
A: Another approach: We have:
$e^x-1=2^x$
If you do not want to use logarithm you can use expansion of $e^x $ and $2^x$:
$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdot\cdot\cdot$
$2^x=(1+1)^x=1+x+\frac{x(x-1)}{2!}\cdot 1^2+\frac{x(x-1)(x_2)}{3!}\cdot 1^3+\frac{x(x-1)(x-2)(x-3)}{4!}\cdot 1^4+\cdot\cdot\cdot$
Taking term with higher powers on both sides give more accurate result; for example taking power 4 on both sides we finally get following equation:
$6x^3+x^2+10x-24+0$
where $x=1.188$ almost satisfies it. However the solution of theses equations is not simple and numeric solutions are much easier.
