Question

simplify.
\\(\left(2w^{-6}x^{5}\
ight)^{-5}\\)
write your answer using only positive exponents.

Explanation:

Step1: Apply the power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). So we apply this to \((2w^{-6}x^{5})^{-5}\), which gives us \(2^{-5}(w^{-6})^{-5}(x^{5})^{-5}\).

Step2: Apply the power of a power rule

The power of a power rule is \((a^m)^n = a^{mn}\). For \(2^{-5}\), it remains as is for now. For \((w^{-6})^{-5}\), we multiply the exponents: \(-6\times(-5)=30\), so it becomes \(w^{30}\). For \((x^{5})^{-5}\), we multiply the exponents: \(5\times(-5)= - 25\), so it becomes \(x^{-25}\). Now our expression is \(2^{-5}w^{30}x^{-25}\).

Step3: Convert negative exponents to positive

Recall that \(a^{-n}=\frac{1}{a^{n}}\) and \(\frac{1}{a^{-n}}=a^{n}\). So \(2^{-5}=\frac{1}{2^{5}}=\frac{1}{32}\) and \(x^{-25}=\frac{1}{x^{25}}\). But we can also rewrite the expression with positive exponents by moving the terms with negative exponents to the appropriate places. So \(2^{-5}w^{30}x^{-25}=\frac{w^{30}}{2^{5}x^{25}}\). Calculating \(2^{5}=32\), we get \(\frac{w^{30}}{32x^{25}}\). Wait, actually, let's do it another way. When we have \((ab^m c^n)^{-p}=a^{-p}b^{-mp}c^{-np}\), and then to make exponents positive, \(a^{-p}=\frac{1}{a^{p}}\), \(b^{-mp}=\frac{1}{b^{mp}}\) if \(mp\) is negative? Wait no, earlier step: \((2w^{-6}x^{5})^{-5}=2^{-5}(w^{-6})^{-5}(x^{5})^{-5}=2^{-5}w^{30}x^{-25}\). Now, \(2^{-5}=\frac{1}{2^{5}}\) and \(x^{-25}=\frac{1}{x^{25}}\), but we can write this as \(\frac{w^{30}}{2^{5}x^{25}}\) or, alternatively, using the rule \((a^m b^n)^{-p}=a^{-p}b^{-np}m^{-p}\)? Wait, no, let's re - apply the exponent rules correctly.

The correct way: \((2w^{-6}x^{5})^{-5}\)

Using \((abc)^n=a^n b^n c^n\), so it's \(2^{-5}(w^{-6})^{-5}(x^{5})^{-5}\)

Now, \((w^{-6})^{-5}=w^{(-6)\times(-5)} = w^{30}\) (power of a power: multiply exponents)

\((x^{5})^{-5}=x^{5\times(-5)}=x^{-25}\)

\(2^{-5}=\frac{1}{2^{5}}=\frac{1}{32}\)

Now, the expression is \(\frac{1}{32}\times w^{30}\times\frac{1}{x^{25}}\) (since \(x^{-25}=\frac{1}{x^{25}}\))

Which is \(\frac{w^{30}}{32x^{25}}\)

Wait, but let's check the exponent rules again. Another approach: \((a^m)^{-n}=a^{-mn}=\frac{1}{a^{mn}}\) when \(mn\) is positive? No, \((a^m)^{-n}=a^{-mn}\), and if we want positive exponents, \(a^{-mn}=\frac{1}{a^{mn}}\) if \(mn>0\), or \(a^{|mn|}\) if \(mn < 0\). Wait, in our case, for the base \(w\): \((w^{-6})^{-5}=w^{(-6)\times(-5)}=w^{30}\) (positive exponent, good). For the base \(x\): \((x^{5})^{-5}=x^{-25}\), so to make it positive, we write \(x^{-25}=\frac{1}{x^{25}}\). For the base 2: \(2^{-5}=\frac{1}{2^{5}}=\frac{1}{32}\). So multiplying them together: \(\frac{1}{32}\times w^{30}\times\frac{1}{x^{25}}=\frac{w^{30}}{32x^{25}}\). Wait, but is there a mistake here? Let's do the exponentiation step by step again.

Original expression: \((2w^{-6}x^{5})^{-5}\)

Using the formula \((k\cdot a^m\cdot b^n)^{-p}=k^{-p}\cdot a^{-mp}\cdot b^{-np}\)

Here, \(k = 2\), \(a = w\), \(m=-6\), \(b = x\), \(n = 5\), \(p = 5\)

So \(k^{-p}=2^{-5}\), \(a^{-mp}=w^{-(-6)\times5}=w^{30}\) (because \(-mp=-(-6)\times5 = 30\)), \(b^{-np}=x^{-5\times5}=x^{-25}\)

So now we have \(2^{-5}\cdot w^{30}\cdot x^{-25}\)

Now, \(2^{-5}=\frac{1}{2^{5}}=\frac{1}{32}\), \(x^{-25}=\frac{1}{x^{25}}\)

So combining these, we get \(\frac{w^{30}}{32x^{25}}\)

Answer:

\(\frac{w^{30}}{32x^{25}}\)