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

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

• A. Convert into a single decimal
• B. Factor the denominator into powers of ten
• C. Multiply by 10 for estimation
• D. Ignore the leading zeroes

Answer: (B)

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

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