# How to express $\log_2 (\sqrt{9} - \sqrt{5})$ in terms of $k=\log_2 (\sqrt{9} + \sqrt{5})$?

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

• Hint: $\log_2(9-5)=\log_2((\sqrt{9}-\sqrt{5})(\sqrt{9}+\sqrt{5}))$ – lemon Aug 24 '14 at 13:23

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

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 $$\log_2 (\sqrt{9} - \sqrt{5}) = 2 - k.$$

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