Special equation?

**Volcanicrain** · May 19th 2008, 01:45 PM

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

$2^x - 5 = x$

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

**Mathstud28** · May 19th 2008, 02:34 PM

Originally Posted by Volcanicrain

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

$2^x - 5 = x$

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

All else fails I always do this

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

so $f(2)=-3$

and $f(4)=7$

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

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

$a_0=2.5$

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

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

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

The check step

$f(3)=0$

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

it works though

**arbolis** · May 19th 2008, 03:43 PM

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

**Mathstud28** · May 19th 2008, 03:50 PM

Originally Posted by arbolis

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

Haha stupid two roots

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