Q. \( [(-\tfrac{1}{2})^{-3}\cdot(-\tfrac{1}{2})^{-1}]^2=[(-\tfrac{1}{2})^{-4}]^2=(-\tfrac{1}{2})^{-8}\).

Q: How do you multiply powers with the same base, e.g. \left(-\frac{1}{2}\right)^{-3}\cdot\left(-\frac{1}{2}\right)^{-1}?

Use a^{m}a^{n}=a^{m+n}. So \left(-\frac{1}{2}\right)^{-3}\cdot\left(-\frac{1}{2}\right)^{-1}=\left(-\frac{1}{2}\right)^{-4}\,.

Q: How do you apply a power to a power, e.g. \left(-\frac{1}{2}\right)^{-4}^{2}?

Use (a^{m})^{n}=a^{mn} . So \left(-\frac{1}{2}\right)^{-4}\right)^{2}=\left(-\frac{1}{2}\right)^{-8}..

Q: How do you divide powers with the same base, e.g. \left(-\frac{1}{2}\right)^{-8} : \left(-\frac{1}{2}\right)^{-3}?

Use a^{m}\div a^{n}=a^{m-n}. Thus \left(-\frac{1}{2}\right)^{-8}\div\left(-\frac{1}{2}\right)^{-3}=\left(-\frac{1}{2}\right)^{-5}\,.

Q: How do you evaluate a negative exponent, e.g. \left(-\frac{1}{2}\right)^{-5}?

a^{-n}=\left(\tfrac{1}{a}\right)^{n} or flip base: \left(-\tfrac{1}{2}\right)^{-5}=\left(-2\right)^{5}=-32. Negative exponent means reciprocal; sign follows parity of exponent..

Q: Why does subtracting a negative become addition, e.g. -\left(-\frac{1}{2}\right)^{-5}?.

Because a-(-b)=a+b. So -(-\frac{1}{2})^{-5} = +(-\frac{1}{2})^{-5}; with values, -(-32)=+32..

Q: How does parity of the exponent affect sign, e.g. \left(-\frac{1}{2}\right)^{n} ?

How does parity of the exponent affect sign, e.g. \left(-\frac{1}{2}\right)^{n} ?

Q: Could a similar expression ever be undefined or infinite?

Only if you try exponent 0 in the denominator, like dividing by 0 or base 0 with negative exponent. Here base -\tfrac{1}{2} is nonzero, so all negative exponents are well-defined.

Answer

\( [(-\tfrac{1}{2})^{-3}\cdot(-\tfrac{1}{2})^{-1}]^2=[(-\tfrac{1}{2})^{-4}]^2=(-\tfrac{1}{2})^{-8}\).

\((- \tfrac{1}{2})^{-8}:(- \tfrac{1}{2})^{-3}=(-\tfrac{1}{2})^{-5},\) and \( (-\tfrac{1}{2})^{-5}=-32,\) so \(-32+32=0.\)

Final result: \(0\)

Detailed Explanation

We simplify the expression step by step:

\( [(-\frac{1}{2})^{-3} \cdot (-\frac{1}{2})^{-1}]^2 : (-\frac{1}{2})^{-3} – (-\frac{1}{2})^{-5} = [(-\frac{1}{2})^{-4}]^2 : (-\frac{1}{2})^{-3} – (-32) = (-\frac{1}{2})^{-8} : (-\frac{1}{2})^{-3} + 32 = (-\frac{1}{2})^{-5} + 32 = -32 + 32 = 0.\)

The original expression is:

\(\left[\left(-\frac{1}{2}\right)^{-3}\cdot\left(-\frac{1}{2}\right)^{-1}\right]^2:\left(-\frac{1}{2}\right)^{-3}-\left(-\frac{1}{2}\right)^{-5}\)

Simplify the product inside the brackets using the rule \(a^m\cdot a^n=a^{m+n}\).\(\left(-\frac{1}{2}\right)^{-3}\cdot\left(-\frac{1}{2}\right)^{-1}=\left(-\frac{1}{2}\right)^{-3+(-1)}\)\(\left(-\frac{1}{2}\right)^{-3+(-1)}=\left(-\frac{1}{2}\right)^{-4}\)
Raise the result to the second power using the rule \((a^m)^n=a^{mn}\).\(\left[\left(-\frac{1}{2}\right)^{-4}\right]^2=\left(-\frac{1}{2}\right)^{-4\cdot2}\)\(\left(-\frac{1}{2}\right)^{-4\cdot2}=\left(-\frac{1}{2}\right)^{-8}\)
Divide by \(\left(-\frac{1}{2}\right)^{-3}\) using the rule \(\frac{a^m}{a^n}=a^{m-n}\).\(\left(-\frac{1}{2}\right)^{-8}:\left(-\frac{1}{2}\right)^{-3}=\left(-\frac{1}{2}\right)^{-8-(-3)}\)\(\left(-\frac{1}{2}\right)^{-8-(-3)}=\left(-\frac{1}{2}\right)^{-5}\)
Now subtract the last term of the original expression.\(\left(-\frac{1}{2}\right)^{-5}-\left(-\frac{1}{2}\right)^{-5}=0\)
A quick numerical check confirms the result.\(\left(-\frac{1}{2}\right)^{-5}=(-2)^5=-32\)\(-32-(-32)=0\)

Final answer: \(0\)

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Arithmetic FAQs

How do you multiply powers with the same base, e.g. \(\left(-\frac{1}{2}\right)^{-3}\cdot\left(-\frac{1}{2}\right)^{-1}\)?

Use \(a^{m}a^{n}=a^{m+n}\). So \(\left(-\frac{1}{2}\right)^{-3}\cdot\left(-\frac{1}{2}\right)^{-1}=\left(-\frac{1}{2}\right)^{-4}\,.\)

How do you apply a power to a power, e.g. \(\left(-\frac{1}{2}\right)^{-4}\)^{2}?

Use \( (a^{m})^{n}=a^{mn} \). So \(\left(-\frac{1}{2}\right)^{-4}\right)^{2}=\left(-\frac{1}{2}\right)^{-8}\)..

How do you divide powers with the same base, e.g. \(\left(-\frac{1}{2}\right)^{-8} : \left(-\frac{1}{2}\right)^{-3}\)?

Use \(a^{m}\div a^{n}=a^{m-n}\). Thus \(\left(-\frac{1}{2}\right)^{-8}\div\left(-\frac{1}{2}\right)^{-3}=\left(-\frac{1}{2}\right)^{-5}\,.\)

How do you evaluate a negative exponent, e.g. \(\left(-\frac{1}{2}\right)^{-5}\)?

\(a^{-n}=\left(\tfrac{1}{a}\right)^{n}\) or flip base: \(\left(-\tfrac{1}{2}\right)^{-5}=\left(-2\right)^{5}=-32\). Negative exponent means reciprocal; sign follows parity of exponent..

Why does subtracting a negative become addition, e.g. \(-\left(-\frac{1}{2}\right)^{-5}\)?.

Because \(a-(-b)=a+b\). So \(-(-\frac{1}{2})^{-5} = +(-\frac{1}{2})^{-5}\); with values, \(-(-32)=+32\)..

How does parity of the exponent affect sign, e.g. \( \left(-\frac{1}{2}\right)^{n} \)?

Why can you simplify stepwise instead of computing decimals immediately?

Laws of exponents keep expressions exact and simpler: combine exponents, then evaluate integer powers. This avoids rounding and reveals cancellations like \(-32+32=0\)..

Could a similar expression ever be undefined or infinite?

Only if you try exponent \(0\) in the denominator, like dividing by \(0\) or base \(0\) with negative exponent. Here base \(-\tfrac{1}{2}\) is nonzero, so all negative exponents are well-defined.

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