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.  

You have an error in one of the steps it should be as follows:
$$x2^x=10 \implies \ln(x2^x)=\ln 10 \implies \ln x+ x \ln 2=\ln 10.$$
To use IVT, let $f(x)=x2^x-10$, now argue that it is a continuous function for $x>0$. After this find two positive real numbers $a$ and $b$ such that $f(a)<0$ and $f(b)>0$. Then by IVT you can claim that there must be some point $c$ in the interval $[a,b]$ where $f$ takes the value $0$. This would mean $c2^c=10$.
If you try $a=1$, you get $f(1)=2-10=-8<0$, now try a value for $b$.
