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.