Find the value of $x$ in this equation. $2^x -x=5$
I can't solve this equation, but I can see that $x=3$ (through trial and error) but I don't know how to attain the value $3$.
 A: There really isn't much you can do here other than trial and error. An equation like $2^x-x = 6$ or $3^x-x = 5$ is completely impossible to solve exactly with any technique taught in school, although I strongly suspect that one can solve it by using, for instance, the Lambert $W$ function in some clever way. This will also let you find the solution to $2^x-x = 5$ that lies close to $-4.97$ (at that point $2^x$ is close to $0$, and $-x$ is close to $5$, so somewhere in that vicinity their sum is exactly $5$).
A: Consider
$$f(x)=2^x -x \implies f'(x)=2^x \log 2-1=0 \implies f''(x)=2^x (\log2)^2>0,$$
therefore $f(x)$ is convex. Since
$$ f'(x)=0 \implies x_0=-\frac{\log \log 2}{\log 2}, \quad f(x_0)=\frac{1}{\log 2}+\frac{\log \log 2}{\log 2}\approx 0.91<5$$
the intermediate value theorem gives us that $f(x)=5$ has exactly $2$ solutions which can be found by numerical methods.
Note that


*

*$f(2)<5<f(4)$

*$f(-4)<5<f(-6)$
therefore the solutions must be in the intervals $(2,4)$ and $(-4,-6)$.
A: $$2^x-x=5$$
Of course, as already said, simple inspection gives the root $x=3$ and the second root close to $-4.968$ can be approach by numerical methods of approximation.
For this second root, there is no exact solution which can be expressed with a finite number of elementary functions. The analytic solution involves a special function, the LambertW function : http://mathworld.wolfram.com/LambertW-Function.html
$$2^{-5}2^{5+x}=5+x$$
$$2^{-5}\ln(2)e^{(5+x)\ln(2)}=(5+x)\ln(2)$$
$$-(5+x)\ln(2)e^{-(5+x)\ln(2)} =-2^{-5}\ln(2)$$
Let $Y=-(5+x)\ln(2) $
$$Ye^Y=X=-2^{-5}\ln(2)$$
With the property of the LambertW function :
$$Y=W(X)=W\big(-2^{-5}\ln(2) \big)$$
$x=-5-\frac{1}{\ln(2)}Y$
$$\boxed{x=-5-\frac{1}{\ln(2)}W\big(-2^{-5}\ln(2) \big)}$$
Note that the LambertW function $W(X)$ is multivalued on the range $-\frac{1}{e}<X<0$
The two branches are noted $W_0(X)$ and $W_{-1}(X)$.
NUMERICALLY :
$X=-2^{-5}\ln(2)\simeq -0.021660849392498$
So we are in the case of multivalued $W(X)$.
First branch : 
$W_0\big(-2^{-5}\ln(2)\big)=-8\ln(2)\simeq -5.545177444479562$
$x\simeq-5-\frac{1}{\ln(2)}(-5.545177444479562)$
$$x\simeq 3.000000000000000$$ 
Second branch :
$W_{-1}(-0.021660849392498)\simeq -0.022145899497751$
$x\simeq-5-\frac{1}{\ln(2)}(-0.022145899497751)$
$$x\simeq -4.96805022061857$$
A: $$2^x - x=5 \implies 2^x  = 5+x$$
It is trivial to see that for $x>3$ , 
$$2^x > 5+x$$
Hence we are only left with $3$ possibilities for $x\in \mathbb {N}$ , which yields $3$ as a solution to the equation.
For $x \in \mathbb {Z}$ , we observe that for $x<-5$,
$$2^x > 5+x$$
Hence the only useful range for the solution is $[-5,3]$ which can be found using trial and error.
