Question

1. $(2b^4)^{-1}$
2. $(2x^4y^{-3})^{-1}$

Explanation:

Problem 17: \(\boldsymbol{(2b^4)^{-1}}\)

Step1: Apply negative exponent rule

The negative exponent rule states that \(a^{-n}=\frac{1}{a^n}\) (or \((ab)^{-n}=\frac{1}{(ab)^n}\) for a product). So, \((2b^4)^{-1}=\frac{1}{2b^4}\) (we can also distribute the exponent: \((2)^{-1}(b^4)^{-1}\), but the result is the same).

Step2: Simplify (if needed)

Using the rule \((a^m)^n = a^{mn}\), \((b^4)^{-1}=b^{-4}=\frac{1}{b^4}\), and \(2^{-1}=\frac{1}{2}\), so multiplying them gives \(\frac{1}{2}\times\frac{1}{b^4}=\frac{1}{2b^4}\).

Step1: Apply negative exponent rule

Using the property \((abc)^{-n}=\frac{1}{(abc)^n}\) (or distributing the exponent: \(2^{-1}(x^4)^{-1}(y^{-3})^{-1}\)). First, distribute the \(-1\) exponent to each factor: \(2^{-1}(x^4)^{-1}(y^{-3})^{-1}\).

Step2: Simplify each factor

• For the coefficient: \(2^{-1}=\frac{1}{2}\)
• For the \(x\) term: \((x^4)^{-1}=x^{4\times(-1)} = x^{-4}=\frac{1}{x^4}\) (or we can keep it as \(x^{-4}\) for now and combine later)
• For the \(y\) term: \((y^{-3})^{-1}=y^{(-3)\times(-1)} = y^{3}\)

Step3: Combine the factors

Now, multiply the simplified factors: \(\frac{1}{2}\times x^{-4}\times y^{3}\). Using the rule \(a^{-n}=\frac{1}{a^n}\), \(x^{-4}=\frac{1}{x^4}\), so we can write this as \(\frac{y^{3}}{2x^{4}}\) (or \(\frac{1}{2}x^{-4}y^{3}\), but the first form is more standard with positive exponents in the numerator/denominator).

Answer:

\(\frac{1}{2b^4}\)

Problem 19: \(\boldsymbol{(2x^4y^{-3})^{-1}}\)