If x = 1/(2^2*4^2*8^2*10^2), how should x be simplified to count non-zero digits?

To simplify the expression ( x = \frac{1}{2^2 \cdot 4^2 \cdot 8^2 \cdot 10^2} ) and find the count of non-zero digits effectively, it is beneficial to factor the denominator into powers of ten.

Firstly, let’s break down each term in the denominator:

• ( 4^2 = (2^2)^2 = 2^4 )

• ( 8^2 = (2^3)^2 = 2^6 )

• ( 10^2 = (2 \cdot 5)^2 = 2^2 \cdot 5^2 )

Next, substituting these factors back into the denominator gives:

[

2^2 \cdot 4^2 \cdot 8^2 \cdot 10^2 = 2^2 \cdot 2^4 \cdot 2^6 \cdot (2^2 \cdot 5^2)

]

Combining the powers of ( 2 ):

[

2^{2 + 4 + 6 + 2} \cdot 5^2