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} |