<PRE><B>Question 701862</B>
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The sum of the logs is the log of the product:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \log(a\ +\ b)\ +\ \log(a\ -\ b)\ =\ \log\left((a\ +\ b)(a\ -\ b)\right)]


You will need to finish the problem by multiplying the two binomials.  Hint:  The product of a pair of conjugates is the difference of two squares.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
&lt;font face=&quot;Math1&quot; size=&quot;+2&quot;&gt;Egw to Beta kai to Sigma&lt;/font&gt;
My calculator said it, I believe it, that settles it
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