Question

question
express the following fraction in simplest form, only using positive exponents.
\\(\frac{10k^{10}}{(2k^{-2})^{-3}}\\)

Explanation:

Step1: Simplify the denominator using exponent rules

Recall the power of a product rule \((ab)^n = a^n b^n\) and the power of a power rule \((a^m)^n=a^{mn}\). For the denominator \((2k^{-2})^{-3}\), we have:
\((2k^{-2})^{-3}=2^{-3}(k^{-2})^{-3}\)
Using the power of a power rule on \((k^{-2})^{-3}\), we get \(k^{(-2)\times(-3)} = k^{6}\). And \(2^{-3}=\frac{1}{2^{3}}=\frac{1}{8}\). So the denominator simplifies to \(\frac{1}{8}k^{6}\). Now the fraction is \(\frac{10k^{10}}{\frac{1}{8}k^{6}}\).

Step2: Divide the coefficients and use exponent rule for division

Dividing by a fraction is multiplying by its reciprocal, so \(\frac{10k^{10}}{\frac{1}{8}k^{6}}=10k^{10}\times\frac{8}{k^{6}}\). Multiply the coefficients: \(10\times8 = 80\). For the variables, use the rule \(\frac{a^m}{a^n}=a^{m - n}\) (here we have multiplication which is like \(a^m\times a^{-n}\) but since we are multiplying by \(\frac{8}{k^{6}}\) it's \(k^{10}\times k^{- 6}\) but actually when dividing \(k^{10}\) by \(k^{6}\) it's \(k^{10-6}\)). So \(k^{10}\div k^{6}=k^{10 - 6}=k^{4}\). Then we have \(80k^{4}\div1\)? Wait, no, wait. Wait, actually, when we have \(\frac{10k^{10}}{\frac{1}{8}k^{6}}\), it's equivalent to \(10\times8\times k^{10}\times k^{6}\)? Wait, no, I made a mistake. Wait, \(\frac{a}{\frac{b}{c}}=\frac{a\times c}{b}\). So \(\frac{10k^{10}}{\frac{1}{8}k^{6}}=\frac{10k^{10}\times8}{k^{6}}\). So \(10\times8 = 80\), and \(\frac{k^{10}}{k^{6}}=k^{10 - 6}=k^{4}\) (using \(\frac{a^m}{a^n}=a^{m - n}\) for \(m>n\)). Wait, but also, let's re - express the original fraction correctly. Alternatively, we can use the rule \(\frac{a^m}{a^n}=a^{m - n}\) and \(\frac{1}{a^{-n}}=a^{n}\). Let's start over.

Original fraction: \(\frac{10k^{10}}{(2k^{-2})^{-3}}\)

First, simplify the denominator \((2k^{-2})^{-3}\). Using \((ab)^n=a^n b^n\), we get \(2^{-3}(k^{-2})^{-3}\). Then \((k^{-2})^{-3}=k^{(-2)\times(-3)} = k^{6}\), and \(2^{-3}=\frac{1}{8}\). So denominator is \(\frac{1}{8}k^{6}\). Now, \(\frac{10k^{10}}{\frac{1}{8}k^{6}}=10k^{10}\times\frac{8}{k^{6}}\) (since dividing by a fraction is multiplying by its reciprocal). Then \(10\times8 = 80\), and for the \(k\) terms, \(\frac{k^{10}}{k^{6}}=k^{10 - 6}=k^{4}\) (using \(\frac{a^m}{a^n}=a^{m - n}\)). Wait, but also, we can use the rule \(\frac{a^m}{a^n}=a^{m - n}\) directly when dividing. Wait, no, the denominator was \((2k^{-2})^{-3}\), let's do it another way. Recall that \(\frac{a^m}{a^n}=a^{m - n}\) and \((a^m)^n=a^{mn}\) and \(a^{-n}=\frac{1}{a^n}\).

Alternative Step1: Use the rule \(\frac{1}{(a^m)^n}=\ a^{-mn}\) or rather, \((a^m)^{-n}=a^{-mn}\). So \((2k^{-2})^{-3}=2^{-3}k^{(-2)\times(-3)}=2^{-3}k^{6}\). Then the fraction is \(\frac{10k^{10}}{2^{-3}k^{6}}\).

Step2: Divide the coefficients and the variables separately. For the coefficients: \(10\div2^{-3}=10\times2^{3}\) (since dividing by a negative exponent is multiplying by the positive exponent). \(2^{3}=8\), so \(10\times8 = 80\). For the variables: \(\frac{k^{10}}{k^{6}}=k^{10 - 6}=k^{4}\) (using \(\frac{a^m}{a^n}=a^{m - n}\)). So putting it together, we get \(80k^{4}\).

Answer:

\(80k^{4}\)