Question 1132319
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            The given condition is not completed.  To be complete,  it must be re-edited in this way:


<pre>
            If a^2+b^2 = 7ab,  <U>where "a" and "b" are positive real numbers</U>,  prove that log{1/3(a+b)}=1/2(log a + log b).
</pre>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;With this re-editing, &nbsp;the solution/(the proof) is as follows.



<pre>
If   a^2 + b^2 = 7ab   then


a^2 + 2ab + b^2 = 9ab


(a + b)^2 = 9ab


log((a+b)^2) = log(9ab)


2*log(a+b) = log(9) + log(a) + log(b)


2*log(a+b) = 2*log(3) + log(a) + log(b)


2*(log(a+b) - log(3)) = log(a) + log(b)


log(a+b) - log(3) = {{{(1/2)*(log((a)) + log((b)))}}}
{{{log(((a+b)/3))}}} = {{{(1/2)*(log((a)) + log((b)))}}}.
</pre>

QED.   &nbsp;&nbsp;The proof is completed.