If the sum of the first 20 terms of the series $$\log_{(7^{1/2})} x + \log_{(7^{1/3})} x + \log_{(7^{1/4})} x + \ldots$$ is 460, then $$x$$ is equal to:

We have the series $$\log_{(7^{1/2})}x+\log_{(7^{1/3})}x+\log_{(7^{1/4})}x+\ldots$$ and we are told that the sum of its first 20 terms equals 460.

First, let us identify the general pattern of the bases. The first term uses the base $$7^{1/2}$$, the second term uses $$7^{1/3}$$, the third uses $$7^{1/4}$$, and so on. Thus the $$n^{\text{th}}$$ term (with $$n$$ starting from 1) uses the base $$7^{1/(n+1)}$$. Therefore, the twentieth term (when $$n=20$$) uses the base $$7^{1/21}$$, confirming that the first 20 terms run from the exponent $$\tfrac12$$ down to $$\tfrac1{21}$$.

Now, we recall the change-of-base formula for logarithms. For any positive $$a,b,c$$ with $$a\neq1$$, the formula is

$$\log_{a^b}c \;=\;\frac{\log_a c}{b}.$$

Applying it to our series where each base is of the form $$7^{1/m},$$ we let $$a=7$$ and $$b=\tfrac1m$$. Hence, for a particular $$m$$ we find

$$\log_{7^{1/m}}x \;=\;\frac{\log_{7}x}{\tfrac1m} \;=\;m\,\log_{7}x.$$

Therefore each term of the given series becomes a simple multiple of $$\log_7 x$$:

$$ \begin{aligned} \log_{(7^{1/2})}x &= 2\,\log_7 x,\\ \log_{(7^{1/3})}x &= 3\,\log_7 x,\\ \log_{(7^{1/4})}x &= 4\,\log_7 x,\\ &\;\;\vdots\\ \log_{(7^{1/21})}x &= 21\,\log_7 x. \end{aligned} $$

Thus the sum of the first 20 terms, which we denote by $$S_{20}$$, is

$$ S_{20} = (2+3+4+\ldots+21)\,\log_7 x. $$

We now need the arithmetic sum $$2+3+4+\ldots+21.$$ Instead of adding each term individually, let us use the well-known formula for the sum of consecutive integers. The sum from 1 to $$n$$ is $$\frac{n(n+1)}2,$$ so

$$1+2+3+\ldots+21=\frac{21\cdot22}2=231.$$

Because we want the sum from 2 to 21, we simply subtract 1:

$$2+3+4+\ldots+21 = 231-1 = 230.$$

Substituting this result into $$S_{20}$$ we obtain

$$S_{20}=230\,\log_7 x.$$

But we are told that $$S_{20}=460.$$ Equating the two expressions gives

$$230\,\log_7 x = 460.$$

Now we isolate $$\log_7 x$$ by dividing both sides by 230:

$$\log_7 x = \frac{460}{230}=2.$$

The definition of the logarithm tells us that if $$\log_7 x = 2,$$ then $$x=7^2.$$ So we find

$$x = 49.$$

This matches Option A, which is $$7^2.$$

Hence, the correct answer is Option A.