Question

If $1,{\log }_9(3^{1-x}+2),{\log }_3({4.3}^x-1)$ are in AP, then x equals

• A. ${\log }_{3}4$
• B. $1-{\log }_{3}4$
• C. $1-{\log }_{4}3$
• D. ${\log }_{4}3$

Answer 1

Answer: (B)

Since, given numbers are in AP.
$\therefore$ $2{\log }_9(3^{1-x}+2)={\log }_3({4.3}^x-1)+1$
$\Rightarrow$ $2{\log }_{3^2}(3^{1-x}+2)={\log }_3(({4.3}^x-1)+{\log }_{3}3$
$\Rightarrow$ $\frac{2}{2}{\log }_3(3^{1-x}+2)={\log }_3[3({4.3}^x-1)]$
$\Rightarrow$ $3^{1-x}+2=3({4.3}^x-1)$
$\Rightarrow$ $\frac{3}{y}+2=12y-3,$ where $y=3^x$
$\Rightarrow$ $12y^2-5y-3=0$
$\Rightarrow$ $y=-\frac{1}{3}$ or $\frac{3}{4}$
$\Rightarrow$ $3^x=-\frac{1}{3}$ or $\frac{3}{4}$
$\Rightarrow$ $x={\log }_3\left(\frac{3}{4}\right)$ $\left(\because 3^x\ne \frac{1}{3}\right)$
$\Rightarrow$ $x=1-{\log }_{3}4$