# How do you simplify fractions that have exponents?

I'm not sure how to simplify a fraction with a large exponent for example:

$$\frac{2^{2001} \cdot 3^{2003}}{6^{2002}}$$

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hint: $6 = 2*3$, and $(2*3)^{2002} = 2^{2002}*3^{2002}$. Are you happy from there? –  user29743 Aug 30 '12 at 22:50
Just a nit-pick about terminology in the title: fractions cannot be solved. They can be simplified. Equations are solved. –  Old John Aug 30 '12 at 23:09

To simplify a fraction with powers in the numerator and denominator a possible method is to factor each power base into prime factors. With practice it can be done directly if the bases are small numbers. $2$ and $3$ are prime numbers. So we need only to factor $6=2\cdot 3$: $$\begin{eqnarray*} \frac{2^{2001}3^{2003}}{6^{2002}} &=&\frac{2^{2001}3^{2003}}{\left( 2\cdot 3\right) ^{2002}}=\frac{2^{2001}3^{2003}}{2^{2002}3^{2002}},\qquad (ab)^n=a^nb^n. \end{eqnarray*}$$

Now we compute the exponents by the rules $\dfrac{a^{p}}{a^{q}}=a^{p-q}$ and $\dfrac{b^{n}}{b^{m}}=\dfrac{1}{b^{m-n}}$ $$\begin{equation*} \frac{2^{2001}3^{2003}}{2^{2002}3^{2002}}=\frac{3^{2003-2003}}{2^{2002-2001}} =\frac{3^{1}}{2^{1}}=\frac{3}{2}, \end{equation*}$$

or factor each power and divide both numerator and denominator by the common factors $$\begin{equation*} \frac{2^{2001}3^{2003}}{2^{2002}3^{2002}}=\frac{2^{2001}3\cdot 3^{2002}}{2\cdot 2^{2001}3^{2002}}=\frac{3}{2}. \end{equation*}$$

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$$\displaystyle \frac{2^{2001} \cdot 3^{2003}}{6^{2002}}= \frac{2^{2001}\cdot 3^{2003}}{2^{2002} \cdot 3^{2002}}=\frac{3}{2}.$$

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$$\frac{2^{2001} 3^{2003}}{6^{2002}}\\ = \frac{2^{2001} 3^{2003}}{2^{2002}3^{2002}}\\ = \frac{2^{2001}}{2^{2002}}\frac{3^{2003}}{3^{2002}}\\ = 2^{2001 - 2002}3^{2003 - 2002} \\= 2^{-1}3^1 = \frac{3}{2}$$

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