# Thread: Prove x=2 if 2^x=x+2;

1. ## Prove x=2 if 2^x=x+2;

If 2^x=x+2. solve it and prove x=2.

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2. ## Re: Prove x=2 if 2^x=x+2;

Originally Posted by GalibMahmud
If 2^x=x+2. solve it and prove x=2.

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If 2^x=x+2. solve it a\
2 ^ 2 = 4
4 = x + 2
x = 4 - 2
x = 2

Steve

3. ## Re: Prove x=2 if 2^x=x+2;

look epsilon1 number
ε1>0 ,ε1<(any (r) from (R+)) so lim(ε1 +r)=r and f(x)`=lim( (f(x+ε1)-f(x))/ε1 )
and integrity f(x+ε1)=f(x±(ε2 or 0))

4. ## Re: Prove x=2 if 2^x=x+2;

Originally Posted by SGS
If 2^x=x+2. solve it a\
You say "solve it" but you did not solve it!

2 ^ 2 = 4
4 = x + 2
x = 4 - 2
x = 2
Replacing $2^s$ by $2^2$, you are assuming from the first that x= 2.

Steve

5. ## Re: Prove x=2 if 2^x=x+2;

Define $f(x)=2^x-x-2$.

$f'(x)=2^x\ln 2-1$

Find critical points:
$x=\dfrac{-\ln (\ln 2)}{\ln 2} \approx 0.5288$. At $x=0, f'(x)<0$ and at $x=1$, $f'(x)>0$. Since $2^0-0-1 <0$, to see if there is another solution, we take the limit as $x\to -\infty$.

$\displaystyle \lim_{x \to -\infty}(2^x-x-2)$ does not exist, as it tends towards infinity. This indicates that there exists exactly one $x <0$ such that $f (x)=0$. While I am able to prove the existence of another solution, I am not sure what it is exactly. It is close to $x\approx -1.69009$