How do I convince students in high school for which this equation: $2^x=4x$ have only one solution in integers that is $x=4$? I would like to convince my student in high school level using a simple
mathematical way  to solve this equation: $$2^x=4x$$ in $\mathbb{Z}$ which
have only one integer solution  that is $x=4$ .
My question here :How do I convince students in high school for which this equation: $$2^x=4x$$ have only one solution that is $x=4$?
Note : I do not want to use substitution to convince them and by numerical methods can't give us exactly $x=4$
EDIT: I edited the question as it is  very related to the precedent
Thank you for any help.
 A: Hint:
plot the graph of $y=2^x$ and $y=4x$ and shows that the only other solution is between $0$ and $1$.
A: If your intention is to 'convince' and not to prove, I'd draw a graph with the functions $y=2^x$ and $y=4x$. The growth rate of each function should make clear that they intersect only at two points, being the first between $0$ and $1$ (and hence, not being an integer).
A: For positive $n$, we have two growing sequences
$$1,2,4,8,\color{green}{16},32,64,128,256\cdots\\
0,4,8,12,\color{green}{16},20,24,28,32\cdots$$
This shows them that the "curves" cross each other at $16$, and it seems that the first grows faster.
Indeed, taking the ratios of successive terms
$$\frac{2^{n+1}}{2^n}=2>\frac{4(n+1)}{4n}=1+\frac1n.$$
A: First of all, the solution of $2^x=4x$ has to be positive.
We have $$\frac{2^x}{x}=4\tag1$$
and let $f(x)=\frac{2^x}{x}$. Then, we have
$$f'(x)=\frac{2^x(x\ln 2-1)}{x^2}$$
So, we know that $f(x)$ is increasing for $x\gt \frac{1}{\ln 2}$ where $1=\frac{1}{\ln e}\lt\frac{1}{\ln 2}\lt \frac{1}{\ln\sqrt e}=2$.
Since $f(1)=2,f(2)=2,f(3)=8/3,f(4)=4$, there is the only one integer solution $x=4$ for $(1)$.
A: We are arguing about numbers $x\in{\mathbb Z}$. When $x\leq 0$ then $2^x>0\geq 4x$, and a case analysis shows that for $1\leq x\leq3$ one has $2^x<4x$. When $x=4$ then obviously $2^x=4x$. It is therefore sufficient to prove that for $x\geq4$ one has
$$2^x\geq 4x\quad\Longrightarrow\quad 2^{x+1}>4(x+1)\ .$$
But this is immediate:
$$2^{x+1}=2\cdot 2^x\geq 2\cdot 4x\geq 4x+16>4(x+1)\ .$$
A: Solutions can occur only for $x>0$ when both sides have the same sign. Take the derivative of $f(x)=2^x-4x$ and see that $f(4)=0$ and $f'(x)>0$ for $x>4$, so the function $f(x)>0$ for $x>4$ since it is growing, hence, no more solutions for $x>4$. Testing other integers $x=1,2,3$ is easy.
A: $$2^x=4x$$
$$\frac{2^x}{4}=x$$
$$2^{x-2}=x.$$
Assume that $x\in \mathbb{Z}$. Clearly, $x\geq 2$, otherwise RHS is non-integral. But the LHS is convex while RHS is linear, hence there can only be $2$ solutions, one for which $\frac{d}{dx}2^{x-2}<\frac{d}{dx} x$, which must occur first, and one for which $\frac{d}{dx} 2^{x-2} > \frac{d}{dx} x$, which occurs later. Therefore, since $\frac{d}{dx} 2^{x-2} > \frac{d}{dx} x$ at $x=4$, this must be the only possible integral solution.
