To solve the given equation
[tex]\[
\left(\frac{1}{4}\right)^{x+1} = 32,
\][/tex]
we'll proceed step-by-step as follows:
1. Write both sides of the equation with the same base:
Recall that [tex]\(\frac{1}{4}\)[/tex] can be written as [tex]\(4^{-1}\)[/tex] and [tex]\(32\)[/tex] can be written as [tex]\(2^5\)[/tex].
So the equation becomes:
[tex]\[
(4^{-1})^{x+1} = 32
\][/tex]
2. Simplify the left side:
Applying the power rule [tex]\((a^m)^n = a^{mn}\)[/tex], we get:
[tex]\[
4^{-(x+1)} = 32
\][/tex]
3. Express 32 as a power of 4:
To further simplify, note that 32 can be written as [tex]\(2^5\)[/tex]. Now we need to express [tex]\(2^5\)[/tex] in terms of a base that involves 4. Remember [tex]\(4 = 2^2\)[/tex], and therefore:
[tex]\[
32 = 2^5 = (2^2)^{2.5} = 4^{2.5}
\][/tex]
4. Set the exponents equal to each other:
Since [tex]\(4^{-(x+1)} = 4^{2.5}\)[/tex], the exponents must be equal:
[tex]\[
-(x + 1) = 2.5
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
Solve the equation [tex]\(-(x + 1) = 2.5\)[/tex]:
[tex]\[
-x - 1 = 2.5
\][/tex]
Adding 1 to both sides:
[tex]\[
-x = 3.5
\][/tex]
Multiplying both sides by -1:
[tex]\[
x = -3.5
\][/tex]
Therefore, the solution to the equation [tex]\(\left(\frac{1}{4}\right)^{x+1} = 32\)[/tex] is
[tex]\[
\boxed{-\frac{7}{2}}
\][/tex]
which corresponds to option B.
