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The question states " Simplify: $$\frac{6\left ( 27^{2n+3} \right )}{9^{3n+6}}$$

What I did was factoring $9$ and and $27$ to make it $3$. In the end I got a $\frac{6}{27}$ answer, simplified: $\frac{2}{9}$. but my problem is that in the book, the answer is stated as $\frac{27}{2}$. What could be the solution?

Shabbeh
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bie
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    Perhaps you could show your work so we can see where your mistake was. – JimmyK4542 Aug 16 '14 at 06:32
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    Either the answer in the book is wrong or perhaps you mistyped the question while copying from the book? – JiK Aug 16 '14 at 06:41
  • Special thanks to all who helped me. Ill be taking a college entrance exam tomorrow, wish me luck. – bie Aug 16 '14 at 06:54

2 Answers2

2

$$\frac{6 \cdot27^{2n+3}}{9^{3n+6}}=\frac{2\cdot3\cdot[(3^3)]^{2n+3}}{(3^2)^{3n+6}}$$

$$=\frac{2\cdot3\cdot3^{6n+9}}{3^{6n+12}}=\frac{2\cdot3}{3^3}=\frac2{3^2}$$

lab bhattacharjee
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$$27^{2n+3}=(3^3)^{2n+3}=3^{6n+9}$$ and $$9^{3n+6}=(3^2)^{3n+6}=3^{6n+12}$$

Mikasa
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