Question

If ${\log }_{4}A={\log }_{6}B={\log }_9(A+B),$ then $\left[\frac{4B}{A}\right]$ ($[\cdot ]$ represents the greatest integer function) is equal to

Answer 1

${\log }_{4}A={\log }_{6}B={\log }_9(A+B)=k\left(say\right)$
$\Rightarrow A=4^k,B=6^k$
and $A+B=9^k\Rightarrow 4^k+6^k=9^k$
$\Rightarrow {\left(\frac{2}{3}\right)}^k+1={\left(\frac{3}{2}\right)}^k$
Let ${\left(\frac{2}{3}\right)}^k=x$
$\Rightarrow x+1=\frac{1}{x}$
$\Rightarrow x^2+x-1=0$
$\Rightarrow x=\frac{1\pm \sqrt{5}}{2}$ but $x>0$
$\Rightarrow x=\frac{1+\sqrt{5}}{2}$
$\therefore \left[4\left(\frac{B}{A}\right)\right]=\left[4x\right]=[2+2\sqrt{5}]=6$