Question

1. an expression is shown.\\(2^{4x - 1}\\)\

which expression is equivalent to the given expression?\
\\(\frac{8^x}{2}\\)\
\\(\frac{16^x}{2}\\)\
\\(8^x - 1\\)\
\\(16^x - 1\\)

Explanation:

Step1: Rewrite base 2 as power of 16

We know that \(2^4 = 16\), so we can rewrite \(2^{4x - 1}\) using the property of exponents \(a^{m+n}=a^m\times a^n\) (or in this case, \(a^{m - n}=\frac{a^m}{a^n}\)). First, rewrite \(2^{4x-1}\) as \(\frac{2^{4x}}{2^1}\). Since \(2^{4x}=(2^4)^x = 16^x\), the expression becomes \(\frac{16^x}{2}\).

Step2: Check other options (optional but for clarity)

• For \(\frac{8^x}{2}\), since \(8 = 2^3\), \(\frac{8^x}{2}=\frac{2^{3x}}{2}=2^{3x - 1}\), which is not equal to \(2^{4x - 1}\) (unless \(x = 0\) or \(x = 1\) in specific cases, but generally not equivalent).
• For \(8^x-1=(2^3)^x-1 = 2^{3x}-1\), which is not of the form \(2^{4x - 1}\) and is a difference, not a quotient of exponents, so not equivalent.
• For \(16^x - 1=(2^4)^x-1=2^{4x}-1\), which is not equal to \(2^{4x - 1}\) as the constant term is different.

Answer:

\(\frac{16^x}{2}\) (the second option, e.g., if options are labeled as A. \(\frac{8^x}{2}\), B. \(\frac{16^x}{2}\), C. \(8^x - 1\), D. \(16^x - 1\), then B. \(\frac{16^x}{2}\))