Logarithms equation, litteral I have problems with the following logarimthic equation:
$$\log _a \left(\frac{x+\sqrt{x^2+5}}{5}\right) = b$$
How can I compute $ \log _a (x-\sqrt{x^2-5})$ in terms of $b$?
 A: I assume you have a typo that is preventing others from answering. Let
$$b=\log_a\left(\frac{x+\sqrt{x^2-5}}{5}\right),\qquad \tilde{b}=\log\left(x-\sqrt{x^2-5}\right).$$
Now use the rules 


*

*$\log(u)+\log(v)=\log(uv)$ 

*$(z-w)(z+w)=z^2-w^2$

*$\log_a(1)=0$


in order to add them together:
$$b+\tilde{b}=\log_a\left(\frac{\color{Red}{x}+\color{Blue}{\sqrt{x^2-5}}}{5}\cdot(\color{Red}{x}-\color{Blue}{\sqrt{x^2-5}})\right)$$
$$=\log_a\left(\frac{\color{Red}{x^2}-\color{Blue}{(x^2-5)}}{5}\right)=\log_a(5/5)=0,$$
hence $\tilde{b}=-b$, as you correctly surmised in the comments.
A: There is presumably a typo in the question. But for fun we show that the version with presumed typo is not as awful as it looks. 
Let $w$ be the second logarithm. Then 
$$5a^b=x+\sqrt{x^2+5}\qquad\text{and}\qquad a^w=x-\sqrt{x^2-5}.$$
Take the first equation, bring $x$ to the left-hand side, square. We get
$$25a^{2b} -10a^b x=5.$$
Operate on the second equation in the same way.
We get
$$a^{2w}-2a^wx=-5.$$
From the first equation, multiplying by $a^w$, we get
$$25a^{2b}a^w -10a^ba^w x=5a^w.$$
From the second equation, multiplying by $5a^b$, we get
$$5a^b a^{2w} -10a^ba^w=-25a^b.$$
Subtract. We get
$$25a^{2b}a^w -5a^ba^{2w}=5a^w+25a^b.$$
This is a quadratic in $a^w$. Solve in the usual way. One root may be extraneous. 
