Question

1. simplify to a single power of base x, and then evaluate for $x = -4$.

a) $\frac{(x^{4})^{5}(x^{3})^{4}}{(x^{7})(x^{23})}$
b) $\frac{(3x^{-5})(-2x^{-2})}{-6x^{-4}}$

Explanation:

Part a)

Step1: Expand exponents via power rule

$(x^4)^5 = x^{4 \times 5} = x^{20}$, $(x^3)^4 = x^{3 \times 4} = x^{12}$, $(x^7)(x^{23}) = x^{7+23} = x^{30}$
Expression becomes: $\frac{x^{20} \cdot x^{12}}{x^{30}}$

Step2: Combine numerator exponents

Add exponents for same base: $x^{20+12} = x^{32}$
Expression becomes: $\frac{x^{32}}{x^{30}}$

Step3: Simplify via quotient rule

Subtract exponents: $x^{32-30} = x^2$

Step4: Evaluate at $x=-4$

Substitute $x=-4$: $(-4)^2 = 16$

Part b)

Step1: Multiply numerator terms

Multiply coefficients: $3 \times (-2) = -6$; Combine exponents: $x^{-5} \cdot x^{-2} = x^{-5-2} = x^{-7}$
Numerator becomes: $-6x^{-7}$
Expression becomes: $\frac{-6x^{-7}}{-6x^{-4}}$

Step2: Simplify coefficients and exponents

Cancel $-6$; Subtract exponents: $x^{-7 - (-4)} = x^{-7+4} = x^{-3} = \frac{1}{x^3}$

Step3: Evaluate at $x=-4$

Substitute $x=-4$: $\frac{1}{(-4)^3} = \frac{1}{-64} = -\frac{1}{64}$

Answer:

a) Simplified form: $x^2$; Value at $x=-4$: $16$
b) Simplified form: $x^{-3}$ or $\frac{1}{x^3}$; Value at $x=-4$: $-\frac{1}{64}$