Question

express the following fraction in simplest form, only using positive exponents.\\(\dfrac{12x^{-2}}{(3p^{-1}x^{-5})^{-3}}\\)

Explanation:

Step1: Simplify the denominator using exponent rules

Recall that \((ab)^n = a^n b^n\) and \((a^m)^n=a^{mn}\). For the denominator \((3p^{-1}x^{-5})^{-3}\), we have:
\(3^{-3}(p^{-1})^{-3}(x^{-5})^{-3}\)
Using \((a^m)^n = a^{mn}\), this becomes \(3^{-3}p^{3}x^{15}\)

Step2: Rewrite the fraction with the simplified denominator

The original fraction is \(\frac{12x^{-2}}{3^{-3}p^{3}x^{15}}\). We can rewrite the constants and variables separately. First, handle the constant terms: \(12\) and \(3^{-3}\). Note that \(12=\frac{12}{1}\) and dividing by \(3^{-3}\) is multiplying by \(3^{3}\). So \(12\times3^{3}\). Calculate \(3^{3}=27\), so \(12\times27 = 324\).

For the variable \(x\): we have \(x^{-2}\) in the numerator and \(x^{15}\) in the denominator. Using the rule \(\frac{a^m}{a^n}=a^{m - n}\), we get \(x^{-2-15}=x^{-17}\), but we need positive exponents, so this is \(\frac{1}{x^{17}}\)

For the variable \(p\): there is no \(p\) in the numerator, so it remains in the denominator as \(p^{3}\)

Step3: Combine the constant and variable terms

Putting it all together, the constant part is \(324\), the \(x\) part is \(\frac{1}{x^{17}}\) and the \(p\) part is \(\frac{1}{p^{3}}\)? Wait, no, wait. Wait, let's re - do the fraction correctly.

Wait, \(\frac{12x^{-2}}{3^{-3}p^{3}x^{15}}=12\times3^{3}\times\frac{x^{-2}}{x^{15}}\times\frac{1}{p^{3}}\)

We know that \(a^{-n}=\frac{1}{a^{n}}\) and \(\frac{a^{m}}{a^{n}}=a^{m - n}\)

First, calculate \(12\times3^{3}\): \(3^{3}=27\), \(12\times27 = 324\)

For the \(x\) terms: \(\frac{x^{-2}}{x^{15}}=x^{-2-15}=x^{-17}=\frac{1}{x^{17}}\)

So putting it all together: \(324\times\frac{1}{p^{3}}\times\frac{1}{x^{17}}=\frac{324}{p^{3}x^{17}}\)? Wait, no, that's wrong. Wait, let's start over with the exponent rules for negative exponents in the denominator.

Wait, the original expression is \(\frac{12x^{-2}}{(3p^{-1}x^{-5})^{-3}}\)

First, recall that \(\frac{1}{a^{-n}}=a^{n}\). So when we have a negative exponent in the denominator, we can move it to the numerator with a positive exponent.

So \((3p^{-1}x^{-5})^{-3}=\frac{1}{(3p^{-1}x^{-5})^{3}}\), so the original fraction is \(12x^{-2}\times(3p^{-1}x^{-5})^{3}\)

Now, expand \((3p^{-1}x^{-5})^{3}\) using \((ab)^n=a^n b^n\) and \((a^m)^n=a^{mn}\):

\(3^{3}(p^{-1})^{3}(x^{-5})^{3}=27p^{-3}x^{-15}\)

Now the expression is \(12x^{-2}\times27p^{-3}x^{-15}\)

Multiply the constants: \(12\times27 = 324\)

Multiply the \(x\) terms: \(x^{-2}\times x^{-15}=x^{-2-15}=x^{-17}=\frac{1}{x^{17}}\)

Multiply the \(p\) terms: \(p^{-3}=\frac{1}{p^{3}}\)

Wait, no, this is the mistake. Wait, when we have \(\frac{A}{B^{-n}}=A\times B^{n}\). So the original denominator is \((3p^{-1}x^{-5})^{-3}\), so \(\frac{12x^{-2}}{(3p^{-1}x^{-5})^{-3}}=12x^{-2}\times(3p^{-1}x^{-5})^{3}\)

Now expand \((3p^{-1}x^{-5})^{3}\):

\(3^{3}(p^{-1})^{3}(x^{-5})^{3}=27p^{-3}x^{-15}\) (since \((a^m)^n=a^{mn}\))

Now multiply with the numerator:

\(12\times27\times x^{-2}\times p^{-3}\times x^{-15}\)

Multiply the constants: \(12\times27 = 324\)

Multiply the \(x\) terms: \(x^{-2}\times x^{-15}=x^{-2 + (-15)}=x^{-17}=\frac{1}{x^{17}}\) (using \(a^m\times a^n=a^{m + n}\))

Multiply the \(p\) terms: \(p^{-3}=\frac{1}{p^{3}}\)

Wait, but this gives \(324\times\frac{1}{p^{3}}\times\frac{1}{x^{17}}=\frac{324}{p^{3}x^{17}}\)? No, that can't be. Wait, let's use the correct exponent rule for negative exponents in the denominator.

Alternative approach:

First, simplify the denominator \((3p^{-1}x^{-5})^{-3}\)

Using \((a^m)^{-n}=a^{-mn}\), so \((3p^{-1}x^{-5})^{-3}=3^{-3}(p^{-1})^{-3}(x^{-5})^{-3}=3^{-3}p^{3}x^{15}\) (since \((a^m)^{-n}=a^{-m\times(-n)}=a^{mn}\))

Now the fraction is \(\frac{12x^{-2}}{3^{-3}p^{3}x^{15}}\)

We can rewrite this as \(12\times3^{3}\times\frac{x^{-2}}{x^{15}}\times\frac{1}{p^{3}}\)

Because \(\frac{1}{3^{-3}} = 3^{3}\)

Calculate \(3^{3}=27\), so \(12\times27 = 324\)

For the \(x\) terms: \(\frac{x^{-2}}{x^{15}}=x^{-2-15}=x^{-17}=\frac{1}{x^{17}}\) (using \(\frac{a^m}{a^n}=a^{m - n}\))

So putting it all together: \(324\times\frac{1}{p^{3}}\times\frac{1}{x^{17}}=\frac{324}{p^{3}x^{17}}\)? Wait, no, that's incorrect. Wait, I think I messed up the sign when moving the terms.

Wait, let's start over with the exponent rules:

The original expression is \(\frac{12x^{-2}}{(3p^{-1}x^{-5})^{-3}}\)

First, apply the exponent rule \((a^m)^{-n}=a^{-mn}\) to the denominator:

\((3p^{-1}x^{-5})^{-3}=3^{-3}(p^{-1})^{-3}(x^{-5})^{-3}=3^{-3}p^{3}x^{15}\) (since \((p^{-1})^{-3}=p^{3}\) and \((x^{-5})^{-3}=x^{15}\))

Now, the fraction is \(\frac{12x^{-2}}{3^{-3}p^{3}x^{15}}\)

We can separate the constants and variables:

For the constants: \(\frac{12}{3^{-3}}=12\times3^{3}\) (because \(\frac{a}{b^{-n}}=a\times b^{n}\))

\(3^{3}=27\), so \(12\times27 = 324\)

For the \(x\) variables: \(\frac{x^{-2}}{x^{15}}=x^{-2-15}=x^{-17}=\frac{1}{x^{17}}\) (using \(\frac{a^m}{a^n}=a^{m - n}\))

For the \(p\) variable: there is no \(p\) in the numerator, so it remains in the denominator as \(p^{3}\)

Wait, but this gives \(\frac{324}{p^{3}x^{17}}\), but that seems wrong. Wait, no, let's check with another method. Let's use the rule of negative exponents in the numerator and denominator.

Recall that \(a^{-n}=\frac{1}{a^{n}}\), so \(x^{-2}=\frac{1}{x^{2}}\) and the denominator \((3p^{-1}x^{-5})^{-3}=\frac{1}{(3p^{-1}x^{-5})^{3}}\)

So the original expression is \(\frac{12\times\frac{1}{x^{2}}}{\frac{1}{(3p^{-1}x^{-5})^{3}}}=12\times\frac{1}{x^{2}}\times(3p^{-1}x^{-5})^{3}\)

Now expand \((3p^{-1}x^{-5})^{3}\):

\(3^{3}(p^{-1})^{3}(x^{-5})^{3}=27p^{-3}x^{-15}\)

So now we have \(12\times\frac{1}{x^{2}}\times27p^{-3}x^{-15}\)

Multiply the constants: \(12\times27 = 324\)

Multiply the \(x\) terms: \(\frac{1}{x^{2}}\times x^{-15}=x^{-2-15}=x^{-17}=\frac{1}{x^{17}}\)

Multiply the \(p\) terms: \(p^{-3}=\frac{1}{p^{3}}\)

So combining these, we get \(324\times\frac{1}{p^{3}}\times\frac{1}{x^{17}}=\frac{324}{p^{3}x^{17}}\)? Wait, this is the same result as before, but let's check with actual numbers. Let's take \(p = 1\) and \(x = 1\)

Original expression: \(\frac{12\times1^{-2}}{(3\times1^{-1}\times1^{-5})^{-3}}=\frac{12}{(3)^{-3}}=12\times27 = 324\)

Our simplified expression: \(\frac{324}{1^{3}\times1^{17}}=324\), which matches. Wait, so maybe it is correct. Wait, but let's take \(x = 2\), \(p = 1\)

Original expression: \(\frac{12\times(2)^{-2}}{(3\times1^{-1}\times(2)^{-5})^{-3}}=\frac{12\times\frac{1}{4}}{(3\times\frac{1}{32})^{-3}}=\frac{3}{(\frac{3}{32})^{-3}}=\frac{3}{(\frac{32}{3})^{3}}=\frac{3\times27}{32768}=\frac{81}{32768}\)

Our simplified expression: \(\frac{324}{1^{3}\times(2)^{17}}=\frac{324}{131072}=\frac{81}{32768}\), which matches. So the simplification is correct.

Wait, but let's re - express the steps correctly:

Step1: Simplify the denominator using \((a^m)^n=a^{mn}\) and \((ab)^n=a^n b^n\)

For \((3p^{-1}x^{-5})^{-3}\), we have:
\[ \begin{align*} (3p^{-1}x^{-5})^{-3}&=3^{-3}(p^{-1})^{-3}(x^{-5})^{-3}\\ &=3^{-3}p^{3}x^{15} \end{align*} \]

Step2: Rewrite the fraction

The original fraction \(\frac{12x^{-2}}{3^{-3}p^{3}x^{15}}\) can be rewritten by moving \(3^{-3}\) to the numerator (since \(\frac{1}{3^{-3}} = 3^{3}\)):
\[ 12\times3^{3}\times\frac{x^{-2}}{x^{15}}\times\frac{1}{p^{3}} \]

Step3: Calculate the constant term and simplify the \(x\) term

Calculate \(3^{3}=27\), so \(12\times27 = 324\)

For the \(x\) term, use \(\frac{a^m}{a^n}=a^{m - n}\):
\[ \frac{x^{-2}}{x^{15}}=x^{-2-15}=x^{-17}=\frac{1}{x^{17}} \]

Step4: Combine all terms

Combining the constant, \(x\) and \(p\) terms, we get:
\[ \frac{324}{p^{3}x^{17}} \]

Answer:

\(\boxed{\dfrac{324}{p^{3}x^{17}}}\)