Question

The roots of the equation $2^{x+2}{27}^{x/(x-1)}=9$ are given by

• A. $1-{\log }_{2}3,2$
• B. ${\log }_2\left(\frac{2}{3}\right),1$
• C. $2,-2$
• D. $1-\frac{\log 3}{\log 2}$

Answer 1

The correct option is D $1-\frac{\log 3}{\log 2}$

Given,

$2^{x+2}{27}^{\frac{x}{x-1}}=9$

$2^{x+2}{3}^{\frac{3x}{x-1}}=3^2$

$2^{x+2}=\frac{3^2}{3^{\frac{3x}{x-1}}}$

$2^{x+2}=3^{2-\frac{3x}{x-1}}$

$2^{x+2}=3^{\frac{2(x-1)-3x}{x-1}}$

$2^{x+2}=3^{\frac{2x-2-3x}{x-1}}$

$2^{x+2}=3^{\frac{-x-2}{x-1}}$

taking log on both sides, we have,

$(x+2)\log 2=\frac{-x-2}{x-1}\log 3$

$(x+2)\log 2=\frac{-(x+2)}{x-1}\log 3$

$\log 2=\frac{-1}{x-1}\log 3$

$(x-1)=-\frac{\log 3}{\log 2}$

$\therefore x=1-\frac{\log 3}{\log 2}$