If $a²+b²=7ab$ where a and b are positive then show that $log(1/3(a+b))=1/2(log a +log b)$

Question

Welcome sir, to the content of my question, please help me:

Could you please provide some context for this question, and your thoughts on it, or progress so far? — Old John, Dec 14 '13 at 12:05
Thank you Old John. For the question I tried to find log(a²+b²) without success. — user115735, Dec 14 '13 at 12:06

score 2 · Accepted Answer · answered Dec 14 '13 at 12:31

2

$a^2+b^2=7ab$ follows $(a+b)^2=9ab$ or $a+b=3 (ab)^{\frac{1}{2}}$ or $\frac{1}{3}(a+b)= (ab)^{\frac{1}{2}}.$ Thus $$ \log\frac{1}{3}(a+b)= \log (ab)^{\frac{1}{2}}=\frac{1}{2}(\log a+\log b). $$

answered Dec 14 '13 at 12:31

Leox

8,120

lab bhattacharjee · Answer 2 · 2013-12-14T12:12:03.093

1

HINT:

As $a^2+b^2=(a+b)^2-2ab,$ we have $(a+b)^2=9ab$

Apply Sum of Logarithms formula ($(2)$ of this)

edited Dec 14 '13 at 12:12

answered Dec 14 '13 at 12:06

lab bhattacharjee

274,582

Thank you so much sir. The greatest source of happiness is the ability to be grateful at all times. – user115735 Dec 14 '13 at 12:08
(a+b)²=9ab
log 1/3(a+b)= log 1/3 x 9ab
log1/3(a+b)= .... Struggle! – user115735 Dec 14 '13 at 12:11
@user115735, why struggle? – lab bhattacharjee Dec 14 '13 at 12:12
I find log1/3(a+b)=log3ab but how is log 3ab equal to 1/2 (log a + log b) – user115735 Dec 14 '13 at 12:19

If $a²+b²=7ab$ where a and b are positive then show that $log(1/3(a+b))=1/2(log a +log b)$

2 Answers2