Solution 3.3:3c
From Förberedande kurs i matematik 1
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- | {{ | + | First, we rewrite the number 0.125 as a fraction which we also simplify |
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- | {{ | + | {{Displayed math||<math>0\textrm{.}125 = \frac{125}{1000} = \frac{5\cdot 25}{10^3} = \frac{5\cdot 5\cdot 5}{(2\cdot 5)^3} = \frac{1}{2^3} = 2^{-3}\,\textrm{.}</math>}} |
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+ | Because 0.125 was expressed as a power of 2, the logarithm can be calculated in full | ||
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+ | {{Displayed math||<math>\log_2 0\textrm{.}125 = \log_2 2^{-3} = (-3)\cdot\log_2 2 = (-3)\cdot 1 = -3\,\textrm{.}</math>}} |
Current revision
First, we rewrite the number 0.125 as a fraction which we also simplify
\displaystyle 0\textrm{.}125 = \frac{125}{1000} = \frac{5\cdot 25}{10^3} = \frac{5\cdot 5\cdot 5}{(2\cdot 5)^3} = \frac{1}{2^3} = 2^{-3}\,\textrm{.} |
Because 0.125 was expressed as a power of 2, the logarithm can be calculated in full
\displaystyle \log_2 0\textrm{.}125 = \log_2 2^{-3} = (-3)\cdot\log_2 2 = (-3)\cdot 1 = -3\,\textrm{.} |