Question

express the following fraction in simplest form, only using positive exponents. \\(\frac{(3a^{5}r^{2})^{-1}}{2a^{-5}r^{-2}}\\)

Explanation:

Step1: Apply negative exponent rule

The negative exponent rule states that \((x^m)^n = x^{mn}\) and \(x^{-n}=\frac{1}{x^n}\) (or \(\frac{1}{x^{-n}} = x^n\)). Apply this to the numerator \((3a^{5}r^{2})^{-1}\):

$$ (3a^{5}r^{2})^{-1}=3^{-1}a^{-5}r^{-2} $$

So the expression becomes \(\frac{3^{-1}a^{-5}r^{-2}}{2a^{-5}r^{-2}}\)

Step2: Cancel out common terms

Notice that \(a^{-5}\) and \(r^{-2}\) appear in both the numerator and the denominator. When we divide terms with the same base, we subtract the exponents (\(x^m\div x^n = x^{m - n}\)). For \(a^{-5}\div a^{-5}\), we have \(a^{-5-(-5)}=a^{0} = 1\) (since any non - zero number to the power of 0 is 1). Similarly, for \(r^{-2}\div r^{-2}\), we have \(r^{-2-(-2)}=r^{0}=1\). So after canceling out \(a^{-5}\) and \(r^{-2}\), the expression simplifies to \(\frac{3^{-1}}{2}\)

Step3: Simplify the remaining fraction

Recall that \(3^{-1}=\frac{1}{3}\). So \(\frac{3^{-1}}{2}=\frac{\frac{1}{3}}{2}\). When dividing by a number is the same as multiplying by its reciprocal, so \(\frac{\frac{1}{3}}{2}=\frac{1}{3}\times\frac{1}{2}=\frac{1}{6}\)

Answer:

\(\frac{1}{6}\)