Question

simplify.
$(2m^{4}n^{3})^{-4}(2m^{2}n)^{7}$
$\frac{?}{m^{\square}n^{\square}}$

Explanation:

Step1: Apply the power of a product rule

For \((2m^{4}n^{3})^{-4}\), we use \((ab)^n = a^n b^n\) and \((a^m)^n=a^{mn}\). So \((2m^{4}n^{3})^{-4}=2^{-4}(m^{4})^{-4}(n^{3})^{-4}=2^{-4}m^{-16}n^{-9}\).
For \((2m^{2}n)^{7}\), we use the same rule: \((2m^{2}n)^{7}=2^{7}(m^{2})^{7}n^{7}=2^{7}m^{14}n^{7}\).

Step2: Multiply the two expressions

Now we multiply \(2^{-4}m^{-16}n^{-9}\) and \(2^{7}m^{14}n^{7}\). For the coefficients, we use \(a^m\times a^n = a^{m + n}\), so \(2^{-4}\times2^{7}=2^{-4 + 7}=2^{3}=8\).
For the \(m\) terms: \(m^{-16}\times m^{14}=m^{-16 + 14}=m^{-2}\).
For the \(n\) terms: \(n^{-9}\times n^{7}=n^{-9 + 7}=n^{-2}\).

Step3: Rewrite with positive exponents in the denominator (if needed)

The expression is \(8m^{-2}n^{-2}\), which can be written as \(\frac{8}{m^{2}n^{2}}\).

Answer:

The numerator is \(8\), the exponent of \(m\) is \(2\), and the exponent of \(n\) is \(2\), so the simplified form is \(\frac{8}{m^{2}n^{2}}\) (or in the given format: numerator \(8\), \(m\) exponent \(2\), \(n\) exponent \(2\)).