Ok guys I will redo step by step the entire process. I am assuming you know the rules of exponents and prime factorization
\(\frac{210^2 \times 21^3 \times 49^{\frac{1}{2}}}{30^4 \times 3} = 7^x\)
Numerator
\(210^2 = (3 \times 7 \times 2 \times 5)^2 = 3^2 \times 7^2 \times 2^2 \times 5^2\)
\(21^3 = (3 \times 7)^3 = 3^3 \times 7^3\)
\(49^{\frac{1}{2}} = \sqrt{49} = 7\)
Denominator
\(30^4 \times 3= 3^4 \times 2^4 \times 5^4 \times 3\)
So
\(\frac{3^2 \times 7^2 \times 2^2 \times 5^2 \times 3^3 \times 7^3 \times 7 }{3^4 \times 2^4 \times 5^4 \times 3}\)
From this , grouping all the common terms we do have
\(\frac{3^5 \times 7^6 \times 2^2 \times 5^2}{3^5 \times 5^4 \times 2^4}\)
The terms canceled out
\(\frac{7^6}{5^2\times 2^2} = \frac{7^6}{10^2}\)
At this point the question is flawed..... there is no way to solve it.
For those who are wondering to check the book, I have already done.
Clearly, the question is badly conceived and written.
Regards
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