Question

The number of integral solution(s) of ${\log }_2(4\times 3^x-6)-{\log }_2(9^x-6)\ge 1$ is

Answer 1

Given: ${\log }_2(4\times 3^x-6)-{\log }_2(9^x-6)\ge 1$
For logarithm to be defined:
$4\times 3^x-6>0,9^x-6>0$
$\Rightarrow 3^x>\frac{3}{2},9^x>6$

Now, ${\log }_2(4\times 3^x-6)-{\log }_2(9^x-6)\ge 1$
$\Rightarrow {\log }_2\left(\frac{4\times 3^x-6}{9^x-6}\right)\ge 1$
$\Rightarrow \frac{4\times 3^x-6}{9^x-6}\ge 2$

Assuming $y=3^x$
$\Rightarrow \frac{4y-6}{y^2-6}\ge 2$
$\Rightarrow 4y-6>2(y^2-6)[9^x>6\Rightarrow y^2>6]$
$\Rightarrow 2y-3\ge y^2-6$
$\Rightarrow y^2-2y-3\le 0$
$\Rightarrow (y+1)(y-3)\le 0$
$\Rightarrow -1\le y\le 3$
$\because 3^x>\frac{3}{2}\Rightarrow y>\frac{3}{2}$
$\therefore \frac{3}{2}<y\le 3$
$\Rightarrow \frac{3}{2}<3^x\le 3$

Hence, the only possible integral solution is
$3^x=3$ i.e, $x=1$