1. ## Special equation?

Wrong forum probably, but I'm not sure where to post...
My friend challenged me with this curious equation:

$\displaystyle 2^x - 5 = x$

Like I know the answer is 3, but I don't know how to mathematically prove this.

2. Originally Posted by Volcanicrain
Wrong forum probably, but I'm not sure where to post...
My friend challenged me with this curious equation:

$\displaystyle 2^x - 5 = x$

Like I know the answer is 3, but I don't know how to mathematically prove this.
All else fails I always do this

$\displaystyle f(x)\equiv{2^x-5-x}$

so $\displaystyle f(2)=-3$

and $\displaystyle f(4)=7$

Therefore by the intermediate value theorem there $\displaystyle \exists{c}\in(2,4)\backepsilon{f(c)}=0$

So now utilizing probably three but for the sake of the absence of luck in math

say the starting point for Newton-Raphson method is 2.5

$\displaystyle a_0=2.5$

$\displaystyle a_2=2.5-\frac{f(2.5)}{f'(2.5)}$

$\displaystyle a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)}$

we see that as $\displaystyle n\to\infty$ $\displaystyle a_n\to{3}$

The check step

$\displaystyle f(3)=0$

Sorry that is bad but I cannot think of anything else as of present

it works though

3. My friend challenged me with this curious equation:

2^x - 5 = x
It might looks simple, but it's a real challenge. In fact, if $\displaystyle f(x)=2^x-5-x$, then f has 2 zeros. It means that the solution is not singular. But to prove it... not easy.
$\displaystyle f'(x)=ln(2)e^{xln(2)}=ln(2)e^x-1$. We want to study its sign. So we want to find when it is equal to $\displaystyle 0$. $\displaystyle f'(x)=0 \Leftrightarrow ln(2)e^x=1 \Leftrightarrow x=ln(\frac{1}{ln(2)})=-ln(ln2)$.
Now if $\displaystyle x<-ln(ln2), f'(x)<0$. It means that $\displaystyle f(x)$ is decreasing from negative infinite to $\displaystyle -ln(ln2)$. As when $\displaystyle x>-ln(ln2)$, $\displaystyle f'(x)>0$, $\displaystyle f(x)$ is increasing from $\displaystyle -ln(ln2)$ to positive infinite.
f(0)=-4. As it will increase (strictly), it must cross the x axis on a single point when $\displaystyle x>0$. You can find a similar conclusion calculating for example $\displaystyle f(-10)$. It means that the roots of f are respectively between $\displaystyle [-10,0]$ and $\displaystyle [0,5]$. (5 because $\displaystyle f(5)>0$). Now to find the roots, you can say that there is an easy one to see, when $\displaystyle x=3$. To calculate the other root, I suggest you the bisection method which will work.

4. Originally Posted by arbolis
It might looks simple, but it's a real challenge. In fact, if $\displaystyle f(x)=2^x-5-x$, then f has 2 zeros. It means that the solution is not singular. But to prove it... not easy.
$\displaystyle f'(x)=ln(2)e^{xln(2)}=ln(2)e^x-1$. We want to study its sign. So we want to find when it is equal to $\displaystyle 0$. $\displaystyle f'(x)=0 \Leftrightarrow ln(2)e^x=1 \Leftrightarrow x=ln(\frac{1}{ln(2)})=-ln(ln2)$.
Now if $\displaystyle x<-ln(ln2), f'(x)<0$. It means that $\displaystyle f(x)$ is decreasing from negative infinite to $\displaystyle -ln(ln2)$. As when $\displaystyle x>-ln(ln2)$, $\displaystyle f'(x)>0$, $\displaystyle f(x)$ is increasing from $\displaystyle -ln(ln2)$ to positive infinite.
f(0)=-4. As it will increase (strictly), it must cross the x axis on a single point when $\displaystyle x>0$. You can find a similar conclusion calculating for example $\displaystyle f(-10)$. It means that the roots of f are respectively between $\displaystyle [-10,0]$ and $\displaystyle [0,5]$. (5 because $\displaystyle f(5)>0$). Now to find the roots, you can say that there is an easy one to see, when $\displaystyle x=3$. To calculate the other root, I suggest you the bisection method which will work.
Haha stupid two roots