Solution 3.3:3f
From Förberedande kurs i matematik 1
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} |
