Question

Given that $2^{x+2}{27}^{\frac{x}{x-1}}=9$, if one solution of this equation is $x=1-\frac{\log 3}{\log 2}$, then find another solution.

• A. $x=-2$
• B. $x=-3$
• C. $x=0$
• D. none of these

Answer 1

The correct option is A $x=-2$

Given $2^{x+2}{27}^{\frac{x}{x-1}}=9$
Taking log on both sides, we get
$(x+2)\log 2+\frac{x}{x-1}\log 27=\log 9[\because \log (ab)=\log a+\log b\text{\&}\log a^m=m\log a]$
$(x+2)\log 2+\frac{x}{x-1}3\log 3=2\log 3$
$(x+2)\log 2+\left(\frac{3x}{x-1}\right)\log 3-2\log 3=0$
$(x+2)[\log 2+\frac{\log 3}{x-1}]=0$
$\Rightarrow x=-2$ or $x-1=-\frac{\log 3}{\log 2}$
$\Rightarrow x=-2,1-\frac{\log 3}{\log 2}$