How do we use the Intermediate Value Theorem to show that $2^x = \frac{10}{x}$ for $x>0$. The Intermediate Value Theorem states that over a closed interval $[a,b]$ for line $L$, that there exists a value $c$ in that interval such that $f(c) = L$.  
We know both functions require $x>0$, however this is not a closed interval. 
However, I went ahead on the problem anyway.  I decided to solve for x. 
$2^x = \frac{10}{x}$
$x2^x = 10$
$x^2log2 = log10$
$x^2 = \frac{1}{log2}$
$x = \sqrt{\frac{1}{log2}}$ 
However, when graphing these two functions I found that they both met up at ~$2.236$, which is not equal to $x = \sqrt{\frac{1}{log2}}$ 
I am very new to Calculus and the Intermediate Value theorem and would appreciate a simple explanation.  Thank you.  

The difference $f(x)=2^x-\frac{10}x$ is continuous on $(0,\infty)$. As $f(1)=-8<0$ and $f(10)=1023>0$, the IVT tells us that there exists at least one $c\in[1,20]$ with $f(c)=0$.
