The correct answer is:
[tex]-\frac{7}{2}[/tex]To solve this, we can apply logarithm:
[tex]N=a^x\Leftrightarrow\log _aN=x[/tex]Then we have:
[tex]\begin{gathered} (\frac{1}{4})^{x+1}=32\Leftrightarrow\log _{\frac{1}{4}}32=x+1 \\ \end{gathered}[/tex]Then we can solve:
[tex]\begin{gathered} \log _{\frac{1}{4}}32=-\frac{5}{2}\Rightarrow-\frac{5}{2}=x+1 \\ \\ x=-\frac{5}{2}-1=-\frac{7}{2} \end{gathered}[/tex]