Solution 3.3:3f

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{{NAVCONTENT_START}}
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If we write
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<center> [[Image:3_3_3f.gif]] </center>
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<math>\text{4}</math>
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{{NAVCONTENT_STOP}}
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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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we obtain
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<math>\begin{align}
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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}2^{-4}=2\centerdot \log _{2}2+\left( -4 \right)\centerdot \log _{2}2 \\
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& =2\centerdot 1+\left( -4 \right)\centerdot 1=-2 \\
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\end{align}</math>

Revision as of 14:28, 25 September 2008

If we write \displaystyle \text{4} and \displaystyle \text{16} as


\displaystyle \begin{align} & \text{4}=2\centerdot 2=2^{2} \\ & 16=2\centerdot 8=2\centerdot 2\centerdot 4=2\centerdot 2\centerdot 2\centerdot 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}} \\ & =\log _{2}2^{2}+\log _{2}2^{-4}=2\centerdot \log _{2}2+\left( -4 \right)\centerdot \log _{2}2 \\ & =2\centerdot 1+\left( -4 \right)\centerdot 1=-2 \\ \end{align}

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