Solution 1.3:4b
From Förberedande kurs i matematik 1
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| - | {{  | + | The numbers 9 and 27 can both be written as powers of 3,  | 
| - | <  | + | |
| - | {{  | + | {{Displayed math||<math>\begin{align}  | 
| + | 9 &= 3\cdot 3 = 3^{2}\,,\\[5pt]  | ||
| + | 27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3}\textrm{.}   | ||
| + | \end{align}</math>}}  | ||
| + | |||
| + | Thus, all factors in the expression can be written using a common base and the whole product can be simplified using the power rules  | ||
| + | |||
| + | {{Displayed math||<math>\begin{align}  | ||
| + | 3^{13}\cdot 9^{-3}\cdot 27^{-2} &= 3^{13}\cdot (3^{2})^{-3}\cdot (3^{3})^{-2}\\[3pt]  | ||
| + | &= 3^{13}\cdot 3^{2\cdot (-3)}\cdot 3^{3\cdot (-2)}\\[3pt]  | ||
| + | &= 3^{13}\cdot 3^{-6}\cdot 3^{-6}\\[3pt]  | ||
| + | &= 3^{13-6-6}\\[3pt]  | ||
| + | &= 3^{1}\\[3pt]  | ||
| + | &= 3\,\textrm{.}  | ||
| + | \end{align}</math>}}  | ||
Current revision
The numbers 9 and 27 can both be written as powers of 3,
| \displaystyle \begin{align}
 9 &= 3\cdot 3 = 3^{2}\,,\\[5pt] 27 &= 3\cdot 9 = 3\cdot 3\cdot 3 = 3^{3}\textrm{.} \end{align}  | 
Thus, all factors in the expression can be written using a common base and the whole product can be simplified using the power rules
| \displaystyle \begin{align}
 3^{13}\cdot 9^{-3}\cdot 27^{-2} &= 3^{13}\cdot (3^{2})^{-3}\cdot (3^{3})^{-2}\\[3pt] &= 3^{13}\cdot 3^{2\cdot (-3)}\cdot 3^{3\cdot (-2)}\\[3pt] &= 3^{13}\cdot 3^{-6}\cdot 3^{-6}\\[3pt] &= 3^{13-6-6}\\[3pt] &= 3^{1}\\[3pt] &= 3\,\textrm{.} \end{align}  | 
