The question is wrong

Fixed and now is correct.

Thanks

I am totally blank about the explanation. Where did you get the 5 and 2 canceled from numerator and denominator?

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

Carcass, x value comes out to be ~2.06. Please check the answer.

From above sir