# 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.

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

$$\lg(a^2+b^2)^2=2 \cdot \lg(a^2+b^2)=2 \biggr( \lg(a+ib)+\lg(a-ib) \biggr)$$
one might use $a^2+b^2=c^2\;$... –  draks ... Sep 4 '12 at 14:11