Question

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

Answer 1

${\log }_{4}A={\log }_{6}B={\log }_9(A+B)=x\text{(Let)}\therefore A=4^x,B=6^x,A+B=9^x\Rightarrow \frac{B}{A}=\frac{6^x}{4^x}=(\frac{3}{2}{)}^x\Rightarrow A+B=9^x\Rightarrow 4^x+6^x=9^x\Rightarrow 2^{2x}+2^x{3}^x-3^{2x}=0\Rightarrow (\frac{3}{2}{)}^{2x}-(\frac{3}{2}{)}^x-1=0\text{Let}(\frac{3}{2}{)}^x=t\therefore t^2-t-1=0\Rightarrow t=\frac{1\pm \sqrt{5}}{2}\because t>0\Rightarrow t=\frac{1+\sqrt{5}}{2}\therefore \frac{B}{A}=\frac{1+\sqrt{5}}{2}\therefore [\frac{4B}{A}]=2(1+\sqrt{2})=6$