Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $b$?

share|cite|improve this question
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
up vote 2 down vote accepted

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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.