$$\rm Method$$ $$1$$:  Since all of the exponents are multiples of $4$, we can simplify the problem by taking the fourth root of each number. Evaluating we get $$10^{2}=100$$, $$5^{3}=125$$, and $$2^{6}=64$$. $$64\textless100\textless125$$. So, $$2^{24} \textless$$ $$10^{8}$$ $$\textless{} 5^{12}$$.  $$\rm Method$$ $$2$$:  First, let us make all exponents equal to $$8$$. Then, it will be easy to order the  numbers without doing any computations. $$10^{8}$$ is fine as it is. We can rewrite $$2^{24}$$ as $$(2^{3})^{8}=8^{8}$$. We can rewrite $$5^{12}$$ as $$\left( 5^{\frac{3}{2}}\right)^{8}=\left( \sqrt{125}\right)^{8}$$. We take the eighth root of all of these to get $$10$$, $$8$$,~$\sqrt{125}$. Obviously, $$8\textless10\textless\sqrt{ 125}$$, so $$2^{24}\textless10^{8}\textless5^{12}$$. 
What is the correct ordering of the three numbers, $$10^{8}$$, $$5^{12}$$, and $$2^{24}$$? ($$2010$$ AMC $$8$$ Problem, Question \#$$24$$) 

(A) $$2^{24}$$\textless{} $$10^{8}$$\textless$$5^{12}$$ 
(B) $$2^{24}$$\textless$$5^{12}$$\textless$$10^{8}$$ 
(C) $$5^{12}$$\textless$$2^{24}$$\textless$$10^{8}$$ 
(D) $$10^{8}$$\textless$$5^{12}$$\textless$$2^{24}$$ 
(E) $$10^{8}$$\textless$$2^{24}$$\textless$$5^{12}$$ 