# 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 $b$?

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Any chance it is $\log_a(x-\sqrt{x^2+5})$? – Tapu Nov 28 '11 at 18:11
@Tapu If it is $\log_a(x-\sqrt{x^2+5})$ log function itself is not defined – Ramana Venkata Nov 28 '11 at 18:26
I get an answer $\log _a (x-\sqrt{x^2-5}) = -b$, but I'm not very sure... – Siscia Nov 28 '11 at 18:31
We have $\log_a(\sqrt{x^2+5}-x)=-b$. Since in general $\sqrt{x^2+5}-x\ne x-\sqrt{x^2-5}$, the answer $-b$ cannot be in general right. I suspect a typo in the problem. With the problem as it stands, possibly an answer can be ground out. – André Nicolas Nov 28 '11 at 18:40
@AndréNicolas $\log_a(x-\sqrt{x^2+5})=-b$ so $\log_a(\sqrt{x^2+5}-x)=b$ what's is wrong ??? I just rationalize the log with $(x-\sqrt{x^2+5})$, so I get $\log_a(\frac{1}{x-\sqrt{x^2+5}})=b$ and then $-\log_a(x-\sqrt{x^2+5}) = b$ and so $\log_a(x-\sqrt{x^2+5}) = -b$, right ? – Siscia Nov 28 '11 at 18:48

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.

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You're making a career out of coloring your equations :-) – Asaf Karagila Nov 28 '11 at 18:52

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.

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You should also answer the question in MO. You seem to be very pround, and probably many reseacher would appreciate your help at their questions. – Sharpie Jun 19 at 17:32