# Simplifying a simple fraction with exponent

I am trying to simplify this fraction : $\dfrac{(3^2)(5^4)}{15^3}$

The answer is : $\dfrac{5}{3}$

I am trying to do the following: $\dfrac{3^2}{15^3} \cdot \dfrac{5^4}{15^3}$ so ... $\dfrac{1^{-3}}{3} \cdot \dfrac{1^1}{3}$

But that's not giving me the right answer, what am I doing wrong, please ?

Thanks !

-
You might want to factor the $15$ and distribute the exponents properly... – J. M. Sep 12 '11 at 8:08

## 2 Answers

For starters, $\displaystyle\frac{3^2}{15^3} \cdot \frac{5^4}{15^3} = \frac{(3^2)(5^4)}{(15^3)(15^3)} = \frac{(3^2)(5^4)}{15^6}$, not $\displaystyle\frac{(3^2)(5^4)}{15^3}$, so your first step isn’t right. Next, $\displaystyle\frac{3^2}{15^3}$ isn’t $\displaystyle\frac{1^{-3}}{3}$, and $\displaystyle\frac{5^4}{15^3}$ isn’t $\displaystyle\frac{1^1}{3}$; these mistakes show that you have some fundamental misconception, but I’m not sure exactly what it is.

Go back to the beginning and write out the fraction without any exponents: \begin{align*} \frac{(3^2)(5^4)}{15^3} &= \frac{3\cdot 3\cdot 5\cdot 5\cdot 5\cdot 5}{15\cdot 15\cdot 15}\\&=\frac{3\cdot 3\cdot 5\cdot 5\cdot 5\cdot 5}{(3\cdot 5)(3\cdot 5)(3\cdot 5)}\\&=\frac{3\cdot 3\cdot 5\cdot 5\cdot 5\cdot 5}{3\cdot 3\cdot 3\cdot 5\cdot 5\cdot 5}\\&=\frac33\cdot\frac33\cdot\frac53\cdot\frac55\cdot\frac55\cdot\frac55\\&=1\cdot 1\cdot\frac53\cdot 1\cdot 1\cdot1\\&=\frac53. \end{align*} When you cancel, you’re really just getting rid of factors that are equal to $1$.

I suspect that you’re actually supposed to be learning to manipulate exponents at this point, but that manipulation is just a shortcut for what I did above. You use the law that $(ab)^n = a^nb^n$ to rewrite the denominator, $15^3$, as $(3\cdot 5)^3 = 3^3 \cdot 5^3$, making your fraction $\displaystyle\frac{3^2 \cdot 5^4}{3^3 \cdot 5^3}$; if you write that out in full, you have the fraction on the third line of the displayed expressions above. Then you split it, correctly this time, as $\displaystyle\frac{3^2}{3^3}\cdot\frac{5^4}{5^3}$. Now, finally, you can use the rule that $\displaystyle\frac{a^n}{a^m}=a^{n-m}$ twice to get $3^{2-3} \cdot 5^{4-3} = 3^{-1} \cdot 5^1 = \displaystyle\frac53$.

-

Here it is explicitly. Notice $15^3=(3\cdot 5)^3=3^3\cdot 5^3$. So $$\dfrac{(3^2)(5^4)}{15^3}=\frac{3^2\cdot 5^4}{3^3\cdot 5^3}=\frac{5}{3}.$$

-