Question

ess the following fraction in simplest form, only using nents. \\(\frac{-3(j^{-3}a^4)^{-3}}{3j^4a^8}\\)

Explanation:

Step1: Simplify the numerator's power

Using the power of a product rule \((ab)^n = a^n b^n\) and power of a power rule \((a^m)^n = a^{mn}\), we simplify the numerator:
\(-3(j^{-3}a^{4})^{-3} = -3j^{(-3)\times(-3)}a^{4\times(-3)} = -3j^{9}a^{-12}\)

Step2: Divide by the denominator

Now we divide the simplified numerator by the denominator \(3j^{4}a^{8}\):
\(\frac{-3j^{9}a^{-12}}{3j^{4}a^{8}}\)
First, divide the coefficients: \(\frac{-3}{3} = -1\)
Then, use the quotient rule for exponents \(a^m\div a^n = a^{m - n}\) for \(j\) and \(a\) terms:
For \(j\): \(j^{9}\div j^{4} = j^{9 - 4} = j^{5}\)
For \(a\): \(a^{-12}\div a^{8} = a^{-12 - 8} = a^{-20}=\frac{1}{a^{20}}\) (but we can keep it with negative exponents as per the problem's "only using exponents" which might allow negative exponents)
Combining these, we get: \(-1\times j^{5}\times a^{-20} = -j^{5}a^{-20}\) or \(-\frac{j^{5}}{a^{20}}\) (if positive exponents are preferred for the denominator)

Answer:

\(-j^{5}a^{-20}\) (or \(-\frac{j^{5}}{a^{20}}\))