If the sum of the first 20 terms of the series  ${\log }_{7^{\frac{1}{2}}}x+{\log }_{7^{\frac{1}{3}}}x+{\log }_{7^{\frac{1}{4}}}x+...$  is 460, then $x$ is equal to

We know that ${\log }_{7^{\frac{1}{2}}}x=2{\log }_{7}x$

Hence, the given series can be written as  $\begin{array}{rcl}{\log }_{7^{\frac{1}{2}}}x+{\log }_{7^{\frac{1}{3}}}x+{\log }_{7^{\frac{1}{4}}}x+...& =& {\log }_{7}x\left(2+3+4+...20 terms\right)\\ & =& {\log }_{7}x\left(230\right)\end{array}$

But the above value given as $460$

Therefore, 

 $\begin{array}{rcl}{\log }_{7}x& =& \frac{460}{230}\\ & =& 2\\ x& =& 7^2\\ & =& \boxed{\mathbf{49}}\end{array}$