Solution 3.3:3f

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Current revision (06:56, 2 October 2008) (edit) (undo)
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If we write
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If we write 4 and 16 as
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<math>\text{4}</math>
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and
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<math>\text{16}</math>
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as
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<math>\begin{align}
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& \text{4}=2\centerdot 2=2^{2} \\
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& 16=2\centerdot 8=2\centerdot 2\centerdot 4=2\centerdot 2\centerdot 2\centerdot 2=2^{4} \\
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\end{align}</math>
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{{Displayed math||<math>\begin{align}
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4 &= 2\cdot 2 = 2^2\,,\\[5pt]
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16 &= 2\cdot 8 = 2\cdot 2\cdot 4 = 2\cdot 2\cdot 2\cdot 2 = 2^4\,,
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\end{align}</math>}}
we obtain
we obtain
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\log_2 4 + \log_2\frac{1}{16}
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& \log _{2}4+\log _{2}\frac{1}{16}=\log _{2}2^{2}+\log _{2}\frac{1}{2^{4}} \\
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&= \log_2 2^2 + \log_2\frac{1}{2^4}\\[5pt]
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& =\log _{2}2^{2}+\log _{2}2^{-4}=2\centerdot \log _{2}2+\left( -4 \right)\centerdot \log _{2}2 \\
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&= \log_2 2^2 + \log_2 2^{-4}\\[5pt]
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& =2\centerdot 1+\left( -4 \right)\centerdot 1=-2 \\
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&= 2\cdot\log_2 2 + (-4)\cdot\log_2 2\\[5pt]
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\end{align}</math>
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&= 2\cdot 1 + (-4)\cdot 1\\[5pt]
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&= -2\,\textrm{.}
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\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}

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