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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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    @Siscia: You are taking the logarithm of a negative quantity. It is very likely that the question has a typo. Out of curiosity I gave an answer that tackles the question as written. But it is probably not the answer to the question as it is *supposed* to be.2011-11-28

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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 :-)2011-11-28
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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.2016-06-19