Question

express the following fraction in simplest form, only using positive exponents.
\frac{6x^2}{(2m^4x^5)^4}

Explanation:

Step1: Simplify the denominator's power

First, we apply the power of a product rule \((ab)^n = a^n b^n\) to the denominator \((2m^{4}x^{5})^{4}\). So we get \(2^{4}(m^{4})^{4}(x^{5})^{4}\). Calculating the powers: \(2^{4}=16\), \((m^{4})^{4}=m^{4\times4}=m^{16}\), and \((x^{5})^{4}=x^{5\times4}=x^{20}\). So the denominator becomes \(16m^{16}x^{20}\). Now our fraction is \(\frac{6x^{2}}{16m^{16}x^{20}}\).

Step2: Simplify the coefficients and the \(x\) terms

First, simplify the coefficient \(\frac{6}{16}\). We can divide numerator and denominator by 2, so \(\frac{6\div2}{16\div2}=\frac{3}{8}\). Then, for the \(x\) terms, we use the rule of exponents \(\frac{a^{m}}{a^{n}}=a^{m - n}\) (where \(a
eq0\) and \(m,n\) are real numbers). So for \(x\), we have \(x^{2-20}=x^{-18}\). But we need only positive exponents, and remember that \(a^{-n}=\frac{1}{a^{n}}\), so \(x^{-18}=\frac{1}{x^{18}}\). Combining the coefficient and the \(x\) term with the \(m\) term (which remains as \(m^{16}\) in the denominator), we get \(\frac{3}{8m^{16}x^{18}}\). Wait, actually, let's re - express it correctly. The original fraction after step 1 is \(\frac{6x^{2}}{16m^{16}x^{20}}\). We can rewrite this as \(\frac{6}{16}\times\frac{x^{2}}{x^{20}}\times\frac{1}{m^{16}}\). Simplifying \(\frac{6}{16}=\frac{3}{8}\), \(\frac{x^{2}}{x^{20}} = x^{2-20}=x^{-18}=\frac{1}{x^{18}}\). So multiplying these together: \(\frac{3}{8}\times\frac{1}{m^{16}}\times\frac{1}{x^{18}}=\frac{3}{8m^{16}x^{18}}\). Alternatively, we can also use the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\) and handle the negative exponent by moving it to the denominator.

Answer:

\(\boldsymbol{\frac{3}{8m^{16}x^{18}}}\)