Question

question
express the following fraction in simplest form, only using positive exponents.
\\(\frac{5b^{-10}v^{-10}}{(3bv^{-1})^{-4}}\\)

Explanation:

Step1: Simplify the denominator using power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). For the denominator \((3bv^{-1})^{-4}\), we apply this rule:
\[ (3bv^{-1})^{-4}=3^{-4}b^{-4}(v^{-1})^{-4} \]
Now, simplify \((v^{-1})^{-4}\) using the power of a power rule \((a^m)^n=a^{mn}\):
\[ (v^{-1})^{-4}=v^{(-1)\times(-4)} = v^{4} \]
So the denominator becomes \(3^{-4}b^{-4}v^{4}\).

Step2: Rewrite the original fraction with the simplified denominator

The original fraction is \(\frac{5b^{-10}v^{-10}}{3^{-4}b^{-4}v^{4}}\). When dividing by a fraction, we multiply by its reciprocal, but here we can use the rule for dividing exponents with the same base \(a^m\div a^n=a^{m - n}\).

First, handle the coefficients: \(5\div3^{-4}\). Recall that \(a^{-n}=\frac{1}{a^n}\), so \(3^{-4}=\frac{1}{3^4}=\frac{1}{81}\), and \(5\div\frac{1}{81}=5\times81 = 405\).

Now, handle the \(b\) terms: \(b^{-10}\div b^{-4}=b^{-10-(-4)}=b^{-10 + 4}=b^{-6}\). Using the negative exponent rule \(a^{-n}=\frac{1}{a^n}\), this becomes \(\frac{1}{b^{6}}\). Wait, no, actually, when we have \(\frac{b^m}{b^n}=b^{m - n}\), so \(\frac{b^{-10}}{b^{-4}}=b^{-10-(-4)}=b^{-6}=\frac{1}{b^{6}}\)? Wait, no, let's correct that. Wait, the original fraction is \(\frac{5b^{-10}v^{-10}}{3^{-4}b^{-4}v^{4}}\), so it's \(5\times3^{4}\times b^{-10-(-4)}\times v^{-10 - 4}\). Wait, I made a mistake in step 2. Let's redo step 2.

Actually, the denominator is \((3bv^{-1})^{-4}\), so the original expression is \(5b^{-10}v^{-10}\times(3bv^{-1})^{4}\) (because \(\frac{1}{x^{-n}}=x^{n}\)). Ah, that's a better approach. Using the rule \(\frac{a}{b^{-n}}=a\times b^{n}\). So \(\frac{5b^{-10}v^{-10}}{(3bv^{-1})^{-4}}=5b^{-10}v^{-10}\times(3bv^{-1})^{4}\).

Now, apply the power of a product rule to \((3bv^{-1})^{4}\): \(3^{4}b^{4}(v^{-1})^{4}=81b^{4}v^{-4}\).

Now, multiply the terms:

First, the coefficients: \(5\times81 = 405\).

Then, the \(b\) terms: \(b^{-10}\times b^{4}=b^{-10 + 4}=b^{-6}\).

Then, the \(v\) terms: \(v^{-10}\times v^{-4}=v^{-10-4}=v^{-14}\).

So now we have \(405\times b^{-6}\times v^{-14}\times3^{4}\)? Wait, no, \(5\times3^{4}=5\times81 = 405\), \(b^{-10}\times b^{4}=b^{-6}\), \(v^{-10}\times(v^{-1})^{4}=v^{-10}\times v^{-4}=v^{-14}\). Wait, no, \((v^{-1})^{4}=v^{-4}\), so \(v^{-10}\times v^{-4}=v^{-14}\).

Now, we have \(405\times b^{-6}\times v^{-14}\). But we need positive exponents, so use \(a^{-n}=\frac{1}{a^n}\). So \(b^{-6}=\frac{1}{b^{6}}\) and \(v^{-14}=\frac{1}{v^{14}}\)? Wait, no, that can't be right. Wait, let's check the exponent on \(v\) again.

Wait, \((3bv^{-1})^{4}=3^{4}b^{4}(v^{-1})^{4}=81b^{4}v^{-4}\) (since \((v^{-1})^{4}=v^{-4}\)). Then, multiplying by \(v^{-10}\) (from the numerator), we get \(v^{-10}\times v^{-4}=v^{-14}\). And the \(b\) terms: \(b^{-10}\times b^{4}=b^{-6}\). The coefficient is \(5\times81 = 405\). So now we have \(405b^{-6}v^{-14}\). To convert to positive exponents, we can write this as \(\frac{405}{b^{6}v^{14}}\)? Wait, no, that seems off. Wait, let's go back.

Wait, the original expression:

\[ \frac{5b^{-10}v^{-10}}{(3bv^{-1})^{-4}} \]

First, simplify the denominator: \((3bv^{-1})^{-4}=3^{-4}b^{-4}(v^{-1})^{-4}=3^{-4}b^{-4}v^{4}\) (using \((a^m)^n=a^{mn}\), so \((v^{-1})^{-4}=v^{4}\)).

Now, the expression becomes:

\[ \frac{5b^{-10}v^{-10}}{3^{-4}b^{-4}v^{4}}=5\times3^{4}\times b^{-10-(-4)}\times v^{-10 - 4} \]

Because \(\frac{a^m}{a^n}=a^{m - n}\) and \(\frac{1}{a^{-n}}=a^{n}\). So \(5\div3^{-4}=5\times3^{4}\), \(b^{-10}\div b^{-4}=b^{-10 + 4}=b^{-6}\), \(v^{-10}\div v^{4}=v^{-14}\).

Now, \(3^{4}=81\), so \(5\times81 = 405\). So we have \(405\times b^{-6}\times v^{-14}\). Using the negative exponent rule \(a^{-n}=\frac{1}{a^n}\), this is \(\frac{405}{b^{6}v^{14}}\)? Wait, but that seems to have negative exponents eliminated. But let's check with another approach.

Alternative approach:

First, expand the denominator: \((3bv^{-1})^{-4}=3^{-4}b^{-4}v^{4}\) (since \((ab^c)^{-d}=a^{-d}b^{-cd}c^{-d}\)? Wait, no, \((3bv^{-1})^{-4}=3^{-4}b^{-4}(v^{-1})^{-4}=3^{-4}b^{-4}v^{4}\) (because \((v^{-1})^{-4}=v^{4}\)).

Now, the numerator is \(5b^{-10}v^{-10}\), denominator is \(3^{-4}b^{-4}v^{4}\).

So, dividing the coefficients: \(5\div3^{-4}=5\times3^{4}=5\times81 = 405\).

Dividing the \(b\) terms: \(b^{-10}\div b^{-4}=b^{-10 + 4}=b^{-6}\) (since \(\frac{b^m}{b^n}=b^{m - n}\)).

Dividing the \(v\) terms: \(v^{-10}\div v^{4}=v^{-10 - 4}=v^{-14}\) (since \(\frac{v^m}{v^n}=v^{m - n}\)).

So putting it all together: \(405\times b^{-6}\times v^{-14}=\frac{405}{b^{6}v^{14}}\) (using \(b^{-6}=\frac{1}{b^{6}}\) and \(v^{-14}=\frac{1}{v^{14}}\)).

Wait, but that seems to be the case. Let's verify with exponents:

Original expression:

\[ \frac{5b^{-10}v^{-10}}{(3bv^{-1})^{-4}} = 5b^{-10}v^{-10} \times (3bv^{-1})^{4} \quad (\text{since } \frac{1}{x^{-n}} = x^n) \]

Now, expand \((3bv^{-1})^{4}\):

\[ (3bv^{-1})^{4} = 3^{4}b^{4}(v^{-1})^{4} = 81b^{4}v^{-4} \quad (\text{since } (a^m)^n = a^{mn}, \text{ so } (v^{-1})^{4} = v^{-4}) \]

Now, multiply the numerator terms:

\[ 5b^{-10}v^{-10} \times 81b^{4}v^{-4} = (5 \times 81) \times (b^{-10} \times b^{4}) \times (v^{-10} \times v^{-4}) \]

Calculate each part:

- Coefficients: \(5 \times 81 = 405\)
- \(b\) terms: \(b^{-10 + 4} = b^{-6}\)
- \(v\) terms: \(v^{-10 + (-4)} = v^{-14}\)

So we have \(405b^{-6}v^{-14}\), which is equivalent to \(\frac{405}{b^{6}v^{14}}\) (using \(a^{-n} = \frac{1}{a^n}\)).

Wait, but let's check with specific values. Let \(b = 2\), \(v = 2\).

Original expression:

Numerator: \(5\times(2)^{-10}\times(2)^{-10}=5\times2^{-20}\)

Denominator: \((3\times2\times(2)^{-1})^{-4}=(3\times2^{1 - 1})^{-4}=(3\times2^{0})^{-4}=3^{-4}=\frac{1}{81}\)

So original expression: \(5\times2^{-20}\div\frac{1}{81}=5\times81\times2^{-20}=\frac{405}{2^{20}}\)

Now, the simplified expression: \(\frac{405}{b^{6}v^{14}}\) with \(b = 2\), \(v = 2\): \(\frac{405}{2^{6}\times2^{14}}=\frac{405}{2^{20}}\), which matches. So that's correct.

Answer:

\(\boxed{\dfrac{405}{b^{6}v^{14}}}\)