Question 1043490
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If a^2 + b^2=7 ab, then prove that 2 log (a-b)=log 5+log a+log b
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<pre>
If {{{a^2 + b^2}}} = {{{7ab}}}, then

{{{a^2 - 2ab + b^2}}} = {{{7ab - 2ab}}}    (after adding 2ab to both side)

{{{(a - b)^2}}} = {{{5ab}}}.

Now take the logarithm of both sides (assuming a > b > 0, which MUST be pointed in the condition, but mistakenly missed).
You will get

2*log(a-b) = log(5) + log(a) + log(b).   QED.
</pre>

Solved.