To find the value of [tex]\( x \)[/tex] that solves the equation [tex]\( 2^{-x} - 2 = 4^x - 1 \)[/tex] to the nearest fourth of a unit, we need to evaluate the equation for different values of [tex]\( x \)[/tex] and observe where the left-hand side (LHS) and the right-hand side (RHS) are closest to matching.
First, recall the equation:
[tex]\[ 2^{-x} - 2 = 4^x - 1. \][/tex]
Let's break this down step by step using some strategic test points:
1. When [tex]\( x \approx 0 \)[/tex] (Option A):
- LHS: [tex]\( 2^0 - 2 = 1 - 2 = -1 \)[/tex]
- RHS: [tex]\( 4^0 - 1 = 1 - 1 = 0 \)[/tex]
- Difference: [tex]\(-1 - 0 = -1\)[/tex]
2. When [tex]\( x \approx -0.75 \)[/tex] (Option B):
- This option is expressed somewhat unusually as [tex]\( x^{1/x} = -0.75 \)[/tex]. Evaluating this exactly as written leads to complex solutions, not pertinent here, so let's test [tex]\( x \approx -0.75 \)[/tex]:
- LHS: [tex]\( 2^{0.75} - 2 \approx 1.68179 - 2 = -0.31821 \)[/tex]
- RHS: [tex]\( 4^{-0.75} - 1 = 0.17678 - 1 = -0.82322 \)[/tex]
- Difference: [tex]\(-0.31821 - (-0.82322) \approx 0.50501\)[/tex]
3. When [tex]\( x = -0.50 \)[/tex] (Option C):
- LHS: [tex]\( 2^{0.5} - 2 \approx 1.41421 - 2 = -0.58579 \)[/tex]
- RHS: [tex]\( 4^{-0.5} - 1 \approx 0.5 - 1 = -0.5 \)[/tex]
- Difference: [tex]\(-0.58579 - (-0.5) \approx -0.08579 \)[/tex]
4. When [tex]\( x \approx -1 \)[/tex] (Option D):
- LHS: [tex]\( 2^1 - 2 = 2 - 2 = 0 \)[/tex]
- RHS: [tex]\( 4^{-1} - 1 = 0.25 - 1 = -0.75 \)[/tex]
- Difference: [tex]\(0 - (-0.75) = 0.75\)[/tex]
Upon reviewing these differences, we see that [tex]\( x = -0.50 \)[/tex] provides the smallest difference.
The solution to the equation to the nearest fourth of a unit is:
[tex]\[ \boxed{x = -0.50} \][/tex]
Thus, the correct answer is C.
