Solution 1.3:4a
From Förberedande kurs i matematik 1
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| - | {{ | + | Because the base is the same in both factors, the exponents can be combined according to the power rules |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,</math>.}} |
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| + | Alternatively, the expressions for the powers can be expanded completely and then cancelled out, | ||
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| + | {{Displayed math||<math>\begin{align} | ||
| + | 2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] | ||
| + | &= 2\cdot 2 = 4\,\textrm{.}\end{align}</math>}} | ||
Current revision
Because the base is the same in both factors, the exponents can be combined according to the power rules
| \displaystyle 2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,. |
Alternatively, the expressions for the powers can be expanded completely and then cancelled out,
| \displaystyle \begin{align}
2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] &= 2\cdot 2 = 4\,\textrm{.}\end{align} |
