Question

express the following fraction in simplest form, only using positive exponents.\\(\frac{(3u^{-1})^{-1}}{15ub^5}\\)\
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Explanation:

Step1: Simplify the numerator using exponent rules

Recall that \((ab)^n = a^n b^n\) and \((a^m)^n=a^{mn}\). For \((3u^{-1})^{-1}\), we have:
\((3u^{-1})^{-1}=3^{-1}(u^{-1})^{-1}\)
Using \((a^m)^n = a^{mn}\), \((u^{-1})^{-1}=u^{(-1)\times(-1)} = u^{1}=u\) and \(3^{-1}=\frac{1}{3}\). So the numerator becomes \(\frac{1}{3}u\).

Step2: Rewrite the fraction with the simplified numerator

The original fraction is \(\frac{(3u^{-1})^{-1}}{15ub^{5}}\), substituting the simplified numerator, we get \(\frac{\frac{1}{3}u}{15ub^{5}}\).

Step3: Divide the coefficients and use exponent rules for variables

For the coefficients: \(\frac{\frac{1}{3}}{15}=\frac{1}{3}\times\frac{1}{15}=\frac{1}{45}\).
For the variable \(u\): Using \(\frac{a^m}{a^n}=a^{m - n}\), here \(m = 1\), \(n = 1\), so \(\frac{u}{u}=u^{1-1}=u^{0} = 1\) (since \(a^{0}=1\) for \(a eq0\)).
For the variable \(b\): We have \(b^{0}\) in the numerator (since there is no \(b\) in the numerator) and \(b^{5}\) in the denominator, so \(\frac{1}{b^{5}}\) (or \(b^{-5}\) but we need positive exponents, so we keep it as \(\frac{1}{b^{5}}\)).

Combining these, we have \(\frac{1}{45}\times1\times\frac{1}{b^{5}}=\frac{1}{45b^{5}}\)? Wait, no, wait. Wait, let's re - check step 1 and step 2.

Wait, let's start over with step 1:

Alternative Step 1: Simplify \((3u^{-1})^{-1}\)
Using \((a\times b^m)^n=a^n\times b^{m\times n}\)
\((3u^{-1})^{-1}=3^{-1}\times(u^{-1})^{-1}\)
\(3^{-1}=\frac{1}{3}\), \((u^{-1})^{-1}=u^{(-1)\times(-1)} = u^{1}\)
So numerator is \(\frac{u}{3}\)

Step 2: The fraction is \(\frac{\frac{u}{3}}{15ub^{5}}\)
We can rewrite this as \(\frac{u}{3}\times\frac{1}{15ub^{5}}\)
Now, cancel out the \(u\) terms: \(u\) in the numerator and \(u\) in the denominator cancel (since \(\frac{u}{u} = 1\))
Then we have \(\frac{1}{3}\times\frac{1}{15b^{5}}\)
\(\frac{1}{3\times15b^{5}}=\frac{1}{45b^{5}}\)? Wait, no, that's not right. Wait, maybe I made a mistake in the sign of the exponent.

Wait, let's use the rule \(\frac{a^m}{a^n}=a^{m - n}\) and \((a^m)^n=a^{mn}\) and \(a^{-n}=\frac{1}{a^{n}}\)

Let's simplify the numerator \((3u^{-1})^{-1}\):

\((3u^{-1})^{-1}=3^{-1}\times(u^{-1})^{-1}\) (by \((ab)^n=a^n b^n\))

\(3^{-1}=\frac{1}{3}\), \((u^{-1})^{-1}=u^{(-1)\times(-1)} = u^{1}\) (by \((a^m)^n=a^{mn}\))

So numerator is \(\frac{u}{3}\)

Denominator is \(15ub^{5}\)

Now, \(\frac{\frac{u}{3}}{15ub^{5}}=\frac{u}{3}\div(15ub^{5})=\frac{u}{3}\times\frac{1}{15ub^{5}}\)

Cancel \(u\) from numerator and denominator: \(\frac{1}{3}\times\frac{1}{15b^{5}}\)

Multiply the constants: \(\frac{1}{3\times15}=\frac{1}{45}\)

So the result is \(\frac{1}{45b^{5}}\)? Wait, no, wait, that seems wrong. Wait, maybe I messed up the exponent in the numerator.

Wait, another approach:

First, simplify \((3u^{-1})^{-1}\):

Using \((a^m)^n=a^{m\times n}\) and \((ab)^n=a^n b^n\)

\((3u^{-1})^{-1}=3^{-1}\times(u^{-1})^{-1}=3^{-1}u^{1}\) (since \((-1)\times(-1) = 1\))

\(3^{-1}=\frac{1}{3}\), so numerator is \(\frac{u}{3}\)

Denominator: \(15ub^{5}\)

Now, divide the two: \(\frac{\frac{u}{3}}{15ub^{5}}=\frac{u}{3}\times\frac{1}{15ub^{5}}\)

The \(u\) terms: \(u\) in numerator and \(u\) in denominator: \(\frac{u}{u}=u^{1 - 1}=u^{0}=1\)

The coefficient: \(\frac{1}{3}\times\frac{1}{15}=\frac{1}{45}\)

The \(b\) term: There is no \(b\) in the numerator, so we have \(\frac{1}{b^{5}}\)

So combining, we get \(\frac{1}{45b^{5}}\)? Wait, but let's check with another exponent rule.

Wait, maybe I made a mistake in the numerator simplification. Let's use the rule \((a^m)^{-n}=a^{-mn}\)

\((3u^{-1})^{-1}=3^{-1}u^{(-1)\times(-1)}=3^{-1}u^{1}\), which is the same as before.

Wait, maybe the original problem is \(\frac{(3u^{-1})^{-1}}{15u b^{5}}\), let's write all exponents:

Numerator: \((3u^{-1})^{-1}=3^{-1}u^{1}\)

Denominator: \(15u^{1}b^{5}\)

Now, using \(\frac{a^m}{a^n}=a^{m - n}\) for each base:

For base 3: \(\frac{3^{-1}}{3^{0}}\) (since there is no 3 in the denominator, the exponent of 3 in the denominator is 0) \(=3^{-1-0}=3^{-1}=\frac{1}{3}\)? No, wait, the denominator has 15, which is \(3\times5\), so 15 can be written as \(3^{1}\times5^{1}\)

Let's rewrite the fraction as:

\(\frac{3^{-1}u^{1}}{3^{1}\times5^{1}\times u^{1}\times b^{5}}\)

Now, for base 3: \(\frac{3^{-1}}{3^{1}}=3^{-1 - 1}=3^{-2}=\frac{1}{9}\)

For base u: \(\frac{u^{1}}{u^{1}}=u^{1 - 1}=u^{0}=1\)

For base 5: \(\frac{1}{5^{1}}=5^{-1}=\frac{1}{5}\)

For base b: \(\frac{1}{b^{5}}=b^{-5}\)

Now, multiply the coefficients: \(\frac{1}{9}\times\frac{1}{5}=\frac{1}{45}\)

And the variable terms: \(1\times\frac{1}{b^{5}}\)

So the result is \(\frac{1}{45b^{5}}\)? Wait, that seems correct. Wait, but let's check with actual numbers. Let's take \(u = 2\), \(b = 1\)

Original numerator: \((3\times(2)^{-1})^{-1}=(3\times\frac{1}{2})^{-1}=(\frac{3}{2})^{-1}=\frac{2}{3}\)

Original denominator: \(15\times2\times(1)^{5}=30\)

The fraction is \(\frac{\frac{2}{3}}{30}=\frac{2}{3}\times\frac{1}{30}=\frac{2}{90}=\frac{1}{45}\), and when \(b = 1\), \(\frac{1}{45b^{5}}=\frac{1}{45\times1}=\frac{1}{45}\), which matches.

If \(b = 2\), original numerator: \((3u^{-1})^{-1}\) with \(u = 2\) is \(\frac{2}{3}\)

Denominator: \(15\times2\times(2)^{5}=30\times32 = 960\)

Fraction: \(\frac{\frac{2}{3}}{960}=\frac{2}{3\times960}=\frac{2}{2880}=\frac{1}{1440}\)

\(\frac{1}{45b^{5}}=\frac{1}{45\times32}=\frac{1}{1440}\), which matches. So the simplification is correct.

Answer:

\(\frac{1}{45b^{5}}\)