Question

The roots of the equation $2^{x+2} \cdot 3^{\frac{3 x}{x-1}}=9$ are given by

• A. $\log _{2},\left(\frac{2}{3}\right)-2$
• B. $3,-3$
• C. $-2,1-\frac{\log 3}{\log 2}$
• D. $1-\log _{2} 3,2$

Answer 1

Answer: (C)

We have, $2^{x+2} \cdot 3^{3 x /(x-1)}=9=3^{2}$
$\Rightarrow (x+2) \log 2+\frac{3 x}{x-1} \log 3 =2 \log 3$
$\Rightarrow (x+2) \log 2+\left(\frac{3 x}{x-1}-2\right) \log 3=0$
$\Rightarrow (x+2)\left(\log 2+\frac{1}{x-1} \log 3\right)=0$
$\Rightarrow x=-2 $
or $x=1-\frac{\log 3}{\log 2} .$