Question

2 multiple choice 1 point evaluate: $(-2)^{-5}$ $-\frac{1}{32}$ $\frac{1}{32}$ $10$ $-10$ clear my selection previous next

Explanation:

Step1: Recall negative exponent rule

The negative exponent rule states that \( a^{-n}=\frac{1}{a^{n}} \) for any non - zero real number \( a \) and positive integer \( n \). So, for \( (-2)^{-5} \), we can rewrite it as \( \frac{1}{(-2)^{5}} \).

Step2: Calculate the value of \( (-2)^{5} \)

When we raise a negative number to an odd power, the result is negative. Using the formula \( a^{n}=\underbrace{a\times a\times\cdots\times a}_{n\text{ times}} \), we have \( (-2)^{5}=(-2)\times(-2)\times(-2)\times(-2)\times(-2) \).
\( (-2)\times(-2) = 4 \), \( 4\times(-2)=-8 \), \( -8\times(-2) = 16 \), \( 16\times(-2)=-32 \).

Step3: Find the value of \( \frac{1}{(-2)^{5}} \)

Since \( (-2)^{5}=-32 \), then \( \frac{1}{(-2)^{5}}=\frac{1}{-32}=-\frac{1}{32} \).

Answer:

\( -\frac{1}{32} \) (corresponding to the option \( -\frac{1}{32} \))