To solve the equation [tex]\( \left(\frac{1}{4}\right)^{x+1} = 32 \)[/tex], we can take the following step-by-step approach:
1. Rewrite the equation in terms of exponents:
[tex]\[
\left(\frac{1}{4}\right)^{x+1} = 32
\][/tex]
2. Express the base [tex]\(\frac{1}{4}\)[/tex] as [tex]\((4)^{-1}\)[/tex]:
[tex]\[
\left(4^{-1}\right)^{x+1} = 32
\][/tex]
3. Simplify the left side using the properties of exponents [tex]\((a^m)^n = a^{m \cdot n}\)[/tex]:
[tex]\[
4^{-(x+1)} = 32
\][/tex]
4. Rewrite 32 as a power of 2 for easier manipulation:
[tex]\[
32 = 2^5
\][/tex]
5. Express the base 4 in terms of 2:
[tex]\[
4 = 2^2
\][/tex]
Therefore,
[tex]\[
4^{-(x+1)} = (2^2)^{-(x+1)}
\][/tex]
Which simplifies to:
[tex]\[
2^{-2(x+1)} = 2^5
\][/tex]
6. Since the bases are the same, set the exponents equal to each other:
[tex]\[
-2(x+1) = 5
\][/tex]
7. Solve for [tex]\(x\)[/tex]:
[tex]\[
-2(x+1) = 5
\][/tex]
[tex]\[
-2x - 2 = 5
\][/tex]
[tex]\[
-2x = 5 + 2
\][/tex]
[tex]\[
-2x = 7
\][/tex]
[tex]\[
x = -\frac{7}{2}
\][/tex]
So, the solution to the equation [tex]\( \left(\frac{1}{4}\right)^{x+1} = 32 \)[/tex] is [tex]\( x = -\frac{7}{2} \)[/tex].
The correct answer is B. [tex]\( -\frac{7}{2} \)[/tex].
