Question

perform the indicated operation. assume all variables represent nonzero real numbers.
\\(\frac{5a^{5}\left(a^{-3}\
ight)^{6}}{\left(a^{-3}\
ight)^{-1}}\\)
\\(\frac{5a^{5}\left(a^{-3}\
ight)^{6}}{\left(a^{-3}\
ight)^{-1}} = \square\\)
(simplify your answer. type exponential notation with positive exponents.)

Explanation:

Step1: Simplify the exponents of \(a\) in the numerator and denominator using the power of a power rule \((a^m)^n = a^{mn}\)

For the numerator: \((a^{-3})^6 = a^{-3\times6}=a^{-18}\), so the numerator becomes \(5a^{5}\times a^{-18}\). Using the product rule \(a^m\times a^n = a^{m + n}\), we get \(5a^{5+(-18)}=5a^{-13}\).
For the denominator: \((a^{-3})^{-1}=a^{-3\times(-1)} = a^{3}\).
So now the expression is \(\frac{5a^{-13}}{a^{3}}\).

Step2: Use the quotient rule \(\frac{a^m}{a^n}=a^{m - n}\) to simplify the \(a\) terms

\(\frac{5a^{-13}}{a^{3}}=5a^{-13 - 3}=5a^{-16}\).

Step3: Convert the negative exponent to positive using the rule \(a^{-n}=\frac{1}{a^n}\)

\(5a^{-16}=\frac{5}{a^{16}}\)? Wait, no, wait. Wait, I made a mistake in Step1. Let's redo Step1.

Wait, original expression: \(\frac{5a^{5}(a^{-3})^{6}}{(a^{-3})^{-1}}\)

First, handle the exponents of \(a\) in the numerator and denominator. Let's use the rule for dividing exponents with the same base: \(a^m\div a^n=a^{m - n}\), but also, when we have \((a^m)^n\), it's \(a^{mn}\).

So, let's consider the exponents of \(a\) in the numerator: \(a^{5}\times(a^{-3})^{6}=a^{5}\times a^{-18}=a^{5 + (-18)}=a^{-13}\) (that part was correct).

Denominator: \((a^{-3})^{-1}=a^{3}\) (correct).

But then, the expression is \(5\times\frac{a^{-13}}{a^{3}}=5\times a^{-13 - 3}=5\times a^{-16}\). But we need positive exponents, so \(a^{-16}=\frac{1}{a^{16}}\), so \(5a^{-16}=\frac{5}{a^{16}}\)? Wait, no, that can't be right. Wait, maybe I messed up the exponent operations.

Wait, another approach: When you have \(\frac{(a^m)^n}{(a^m)^p}\), that's \(a^{mn - mp}=a^{m(n - p)}\). Let's let \(m=-3\), \(n = 6\) (from numerator's \((a^{-3})^6\)) and \(p=-1\) (from denominator's \((a^{-3})^{-1}\)). Wait, no, the numerator has \(a^{5}\times(a^{-3})^6\), so the total exponent of \(a\) in the numerator is \(5 + (-3)\times6\), and the exponent in the denominator is \((-3)\times(-1)\)? No, denominator is \((a^{-3})^{-1}\), so exponent of \(a\) in denominator is \((-3)\times(-1)=3\). So numerator exponent: \(5 + (-3)\times6=5 - 18=-13\). Denominator exponent: 3. So the exponent of \(a\) in the fraction is \(-13-3=-16\). So \(5a^{-16}\), which is \(\frac{5}{a^{16}}\)? Wait, but that seems off. Wait, maybe I made a mistake in the sign when subtracting.

Wait, no, let's start over.

Original expression: \(\frac{5a^{5}(a^{-3})^{6}}{(a^{-3})^{-1}}\)

First, apply the power of a power rule to \((a^{-3})^{6}\) and \((a^{-3})^{-1}\):

\((a^{-3})^{6}=a^{-3\times6}=a^{-18}\)

\((a^{-3})^{-1}=a^{-3\times(-1)}=a^{3}\)

So now the expression is \(\frac{5a^{5}\times a^{-18}}{a^{3}}\)

Now, combine the exponents in the numerator: \(a^{5}\times a^{-18}=a^{5 + (-18)}=a^{-13}\) (since \(a^m\times a^n=a^{m + n}\))

Now, we have \(\frac{5a^{-13}}{a^{3}}\)

Now, use the quotient rule: \(\frac{a^m}{a^n}=a^{m - n}\), so \(a^{-13}\div a^{3}=a^{-13 - 3}=a^{-16}\)

So the expression is \(5a^{-16}\), and to write with positive exponents, \(a^{-16}=\frac{1}{a^{16}}\), so \(5a^{-16}=\frac{5}{a^{16}}\)? Wait, but that seems incorrect. Wait, no, wait, maybe I messed up the exponent in the numerator. Wait, the original numerator is \(5a^{5}(a^{-3})^{6}\), so \(a^{5}\times(a^{-3})^{6}=a^{5}\times a^{-18}=a^{5 - 18}=a^{-13}\), correct. Denominator is \((a^{-3})^{-1}=a^{3}\), correct. Then \(\frac{a^{-13}}{a^{3}}=a^{-13 - 3}=a^{-16}\), correct. So \(5a^{-16}=\frac{5}{a^{16}}\)? But that seems odd. Wait, maybe I made a mistake in the exponent rules.

Wait, another way: Let's use the rule for exponents when multiplying and dividing. The expression is \(5a^{5}\times(a^{-3})^{6}\times(a^{-3})^{1}\) (because dividing by \((a^{-3})^{-1}\) is multiplying by \((a^{-3})^{1}\), since \(\frac{1}{(a^{-3})^{-1}}=(a^{-3})^{1}\)). Wait, no: \(\frac{1}{(a^m)^n}=(a^m)^{-n}\), so \(\frac{1}{(a^{-3})^{-1}}=(a^{-3})^{1}\), because \((a^{-3})^{-1}\) in the denominator, so reciprocal is \((a^{-3})^{1}\). So then, the expression becomes \(5a^{5}\times(a^{-3})^{6}\times(a^{-3})^{1}\). Now, combine the exponents of \(a\): \(5 + (-3)\times6+(-3)\times1\)? Wait, no, \(a^{5}\times(a^{-3})^{6}\times(a^{-3})^{1}=a^{5}\times a^{-18}\times a^{-3}=a^{5 - 18 - 3}=a^{-16}\), same as before. So \(5a^{-16}=\frac{5}{a^{16}}\). Wait, but that seems correct? Wait, let's test with a number. Let \(a = 2\).

Original expression: \(\frac{5\times2^{5}\times(2^{-3})^{6}}{(2^{-3})^{-1}}\)

Calculate numerator: \(5\times32\times(1/8)^{6}\). \((1/8)^{6}=1/262144\), so \(5\times32\times1/262144=160/262144=5/8192\).

Denominator: \((2^{-3})^{-1}=2^{3}=8\). Wait, no, original expression is \(\frac{5a^{5}(a^{-3})^{6}}{(a^{-3})^{-1}}\), so with \(a = 2\), numerator: \(5\times2^{5}\times(2^{-3})^{6}=5\times32\times2^{-18}=160\times1/262144=160/262144=5/8192\). Denominator: \((2^{-3})^{-1}=2^{3}=8\). So the expression is \((5/8192)\div8=5/(8192\times8)=5/65536\). Now, \(\frac{5}{a^{16}}\) when \(a = 2\) is \(5/65536\), which matches. So that's correct. So the answer is \(\frac{5}{a^{16}}\)? Wait, but the problem says "type exponential notation with positive exponents". Wait, \(\frac{5}{a^{16}}\) is exponential notation with positive exponent (since \(a^{16}\) has positive exponent). Alternatively, \(5a^{-16}\) has negative exponent, so we need to write it as \(\frac{5}{a^{16}}\) or \(5a^{-16}\) is not allowed, so \(\frac{5}{a^{16}}\). Wait, but let's check the exponent calculation again.

Wait, original expression: \(\frac{5a^{5}(a^{-3})^{6}}{(a^{-3})^{-1}}\)

Using the rule \(\frac{(a^m)^n}{(a^m)^p}=a^{m(n - p)}\), where \(m=-3\), \(n = 6\) (from numerator's \((a^{-3})^6\)) and \(p=-1\) (from denominator's \((a^{-3})^{-1}\)), but also we have \(a^{5}\) in the numerator. So total exponent of \(a\) is \(5 + m(n - p)=5+(-3)(6 - (-1))=5+(-3)(7)=5 - 21=-16\), same as before. So \(5a^{-16}=\frac{5}{a^{16}}\).

Yes, that's correct.

Answer:

\(\boxed{\dfrac{5}{a^{16}}}\)