Solution 3.3:3f
From Förberedande kurs i matematik 1
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| - | {{ | + | If we write 4 and 16 as |
| - | < | + | |
| - | {{ | + | {{Displayed math||<math>\begin{align} |
| + | 4 &= 2\cdot 2 = 2^2\,,\\[5pt] | ||
| + | 16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,, | ||
| + | \end{align}</math>}} | ||
| + | |||
| + | we obtain | ||
| + | |||
| + | {{Displayed math||<math>\begin{align} | ||
| + | \log_2 4 + \log_2\frac{1}{16} | ||
| + | &= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt] | ||
| + | &= \log_2 2^2 + \log_2 2^{-4}\\[5pt] | ||
| + | &= 2\cdot\log_2 2 + (-4)\cdot\log_2 2\\[5pt] | ||
| + | &= 2\cdot 1 + (-4)\cdot 1\\[5pt] | ||
| + | &= -2\,\textrm{.} | ||
| + | \end{align}</math>}} | ||
Current revision
If we write 4 and 16 as
| \displaystyle \begin{align}
4 &= 2\cdot 2 = 2^2\,,\\[5pt] 16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,, \end{align} |
we obtain
| \displaystyle \begin{align}
\log_2 4 + \log_2\frac{1}{16} &= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt] &= \log_2 2^2 + \log_2 2^{-4}\\[5pt] &= 2\cdot\log_2 2 + (-4)\cdot\log_2 2\\[5pt] &= 2\cdot 1 + (-4)\cdot 1\\[5pt] &= -2\,\textrm{.} \end{align} |
