# Use identities to simplify $lg(a^2+b^2)^2$

Use logarithmic identities to simply the following:

$$lg(a^2+b^2)^2$$

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.

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$a^2+b^2 \neq (a+b)(a-b)$ 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

## 1 Answer

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

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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