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If $$k=\log_2 (\sqrt{9} + \sqrt{5})$$ express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k$.

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  • $\begingroup$ Hint: $\log_2(9-5)=\log_2((\sqrt{9}-\sqrt{5})(\sqrt{9}+\sqrt{5}))$ $\endgroup$ – lemon Aug 24 '14 at 13:23
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Adding the logarithms $\log_2{(\sqrt{9}-\sqrt{5})}$ and $\log_2{(\sqrt{9}+\sqrt{5})}$ we get the following:

$$\log_2{(\sqrt{9}-\sqrt{5})}+\log_2{(\sqrt{9}+\sqrt{5})}=\log_2{(\sqrt{9}-\sqrt{5})(\sqrt{9}+\sqrt{5})}=\log_2{(9-5)}=\log_2{4}=\log_2{2^2}=2 \cdot \log_2{2}=2$$

Knowing that $k=\log_2{(\sqrt{9}+\sqrt{5})}$ we have:

$$\log_2{(\sqrt{9}-\sqrt{5})}+\log_2{(\sqrt{9}+\sqrt{5})}=\log_2{(\sqrt{9}-\sqrt{5})}+k$$

Therefore, $$2=\log_2{(\sqrt{9}-\sqrt{5})}+k \Rightarrow \log_2{(\sqrt{9}-\sqrt{5})}=2-k$$

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Add to both sides the term $\log_2 (\sqrt{9} - \sqrt{5})$, then you have $$ \log_2 (\sqrt{9} - \sqrt{5}) + k = \log_2 (4), $$ so that $$ \log_2 (\sqrt{9} - \sqrt{5}) = 2 - k. $$

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    $\begingroup$ Add not multiply. $\endgroup$ – lemon Aug 24 '14 at 13:24
  • $\begingroup$ Yuck, what a blunder. Thanks. $\endgroup$ – Bennett Gardiner Aug 24 '14 at 13:26
  • $\begingroup$ The answer from the text book is 2(2-k) $\endgroup$ – BillyTeo Aug 24 '14 at 13:29
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    $\begingroup$ @BillyTeo well the textbook is wrong. $\endgroup$ – lemon Aug 24 '14 at 13:31

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