The product of all real [roots](https://artofproblemsolving.com/wiki/index.php/Root_(polynomial)) of the equation $x^{\log_{10}{x}}=10$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }-1 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }10^{-1} \qquad \mathrm{(E) \ } \text{None of these}$
Take the [logarithm](https://artofproblemsolving.com/wiki/index.php/Logarithm) base $x$ of both sides to get $\log_{10}x=\log_x10$. Use the [change of base formula](https://artofproblemsolving.com/wiki/index.php/Change_of_base_formula) to get $\log_{10}x=\frac{\log_{10}10}{\log_{10}x}=\frac{1}{\log_{10}x}$. Therefore, $(\log_{10}x)^2=1$ and $\log_{10}x=\pm1$. From $\log_{10}x=1$, we have $x=10^1=10$ and from $\log_{10}x=-1$ we have $x=10^{-1}=\frac{1}{10}$. The product is thus $10\times\frac{1}{10}=1, \boxed{\text{A}}$.