Question

1. $\left(\frac{5}{a^4}\

ight)^{-2}$

1. $(4x^5)^2$
2. $3x^{-6}$
3. $4^{-3}$

Explanation:

Problem 1: $(\frac{5}{9})^{-2}$

Step 1: Apply the negative exponent rule

The negative exponent rule states that \(a^{-n}=\frac{1}{a^{n}}\) or equivalently \((\frac{a}{b})^{-n}=(\frac{b}{a})^{n}\). So for \((\frac{5}{9})^{-2}\), we can rewrite it as \((\frac{9}{5})^{2}\).

Step 2: Square the fraction

Squaring a fraction \(\frac{a}{b}\) gives \(\frac{a^{2}}{b^{2}}\). So \((\frac{9}{5})^{2}=\frac{9^{2}}{5^{2}}=\frac{81}{25}\).

Step 1: Apply the power of a product rule

The power of a product rule is \((ab)^{n}=a^{n}b^{n}\). Here \(a = 4\), \(b=x^{5}\) and \(n = 2\). So we have \(4^{2}\times(x^{5})^{2}\).

Step 2: Simplify each part

First, \(4^{2}=16\). Then, using the power of a power rule \((a^{m})^{n}=a^{mn}\), for \((x^{5})^{2}\), we get \(x^{5\times2}=x^{10}\).

Step 3: Multiply the results

Multiplying \(16\) and \(x^{10}\) gives \(16x^{10}\).

Step 1: Apply the negative exponent rule

Using the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\), we can rewrite \(x^{-6}\) as \(\frac{1}{x^{6}}\).

Step 2: Multiply by the coefficient

So \(3x^{-6}=3\times\frac{1}{x^{6}}=\frac{3}{x^{6}}\). (If the original problem was \(3x^{-6}\times6x\) or something else, the steps would change, but based on the given "18x" and "3x^{-6}", maybe there was a multiplication error. Assuming it's just simplifying \(3x^{-6}\), the above is the solution. If it was \(3x^{-6}\times6x\):

Step 1: Multiply the coefficients and apply the product of powers rule

The product of powers rule is \(a^{m}\times a^{n}=a^{m + n}\). So \(3\times6=18\) and \(x^{-6}\times x=x^{-6 + 1}=x^{-5}=\frac{1}{x^{5}}\). Then the result would be \(18\times\frac{1}{x^{5}}=\frac{18}{x^{5}}\). But based on the given "18x" and "3x^{-6}", perhaps the intended problem was \(3x^{-6}\times6x\), so we'll go with that correction.)

Answer:

\(\frac{81}{25}\)

Problem 2: \((4x^{5})^{2}\)