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

Question

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

Answer 1

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

Answer 2

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