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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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    $a^2+b^2 \neq (a+b)(a-b)$ so $lg(a^2+b^2)\neq lg(a+b)+lg(a-b)$.2012-09-04
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    Arg of course you are right. I edited the mistake.2012-09-04
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    Iuli, post in an answer so the question isn't left unanswered?2012-09-04
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    I'd think $2\log(a^2 + b^2)$ is the natural stopping point. Any particular reason you think you can proceed further?2012-09-04
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    Seems to be too easy. But I also don't see a way to go on, so it should be the stopping point.2012-09-04
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    What do you mean by simplify in this context? More efficient calculation or more elegant form?2012-09-04
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    I mean a more elegant form.2012-09-04
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    What about $4\lg c$, since $a^2+b^2=c^2$?2012-09-04

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

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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.2012-09-04
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    Does introducing complex numbers really count as a "simplifying"?2012-09-04
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    Yes, I think compley is not simplifying. Perhaps any answers from the comments above? I'll mark is as solved then2012-09-04
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    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.2012-09-04
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    one might use $ a^2+b^2=c^2\;$...2012-09-04