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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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3  
$a^2+b^2 \neq (a+b)(a-b)$ 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
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1 Answer

up vote 0 down vote accepted

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

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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
1  
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
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