Question

If $a^2-12ab+4b^2=0$, Find the value of : $-\log \left(a+2b\right)+\frac{1}{2}\left(\log a+\log b\right)+2\log 2$.

Answer 1

$a^2-12ab+4b^2=0$

$⟹a^2-12ab+36b^2-32b^2=0$

$⟹(a-6b{)}^2=(4\sqrt{2}{)}^2$

$⟹a=6b=\pm 4\sqrt{2}b$

$a=4\sqrt{2}b+6b,a=-4\sqrt{2}b+6b$

$\therefore a=\pm 4\sqrt{2}b+6b$

$=-\log (a+2b)+\frac{1}{2}(\log a+\log b)+2\log 2$

$=-\log (\pm 4\sqrt{2}b+8b)+\frac{1}{2}\log \left(b\right(\pm 4\sqrt{2}b+8b\left)\right)+2\log 2$

$=0$