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

Use logarithmic identities to simply the following:


I started with

\begin{eqnarray} lg(a^2+b^2)^2&=&2 \cdot lg(a^2+b^2) \\ \end{eqnarray}

I think it's not the final result, but I don't know how to proceed. Any hints would be helpful.

share|cite|improve this question
$a^2+b^2 \neq (a+b)(a-b)$ so $lg(a^2+b^2)\neq lg(a+b)+lg(a-b)$. – Iuli Sep 4 '12 at 7:18
Arg of course you are right. I edited the mistake. – ulead86 Sep 4 '12 at 7:22
Iuli, post in an answer so the question isn't left unanswered? – Kirk Boyer Sep 4 '12 at 7:23
I'd think $2\log(a^2 + b^2)$ is the natural stopping point. Any particular reason you think you can proceed further? – EuYu Sep 4 '12 at 7:29
Seems to be too easy. But I also don't see a way to go on, so it should be the stopping point. – ulead86 Sep 4 '12 at 7:32
up vote 0 down vote accepted

$$\lg(a^2+b^2)^2=2 \cdot \lg(a^2+b^2)=2 \biggr( \lg(a+ib)+\lg(a-ib) \biggr) $$

share|cite|improve this answer
That's hardly what the OP wanted. Besides you have to be very careful with the choice of branches if you want the last equality to be true. – mrf Sep 4 '12 at 7:39
Does introducing complex numbers really count as a "simplifying"? – Dan Sep 4 '12 at 7:50
Yes, I think compley is not simplifying. Perhaps any answers from the comments above? I'll mark is as solved then – ulead86 Sep 4 '12 at 8:14
I agree with draks. This is the only symplification possible. I would have answered the same. If this is not the right answer id say there is no answer. – mick Sep 4 '12 at 9:25
one might use $ a^2+b^2=c^2\;$... – draks ... Sep 4 '12 at 14:11

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.