Question

If $1,{\log }_3(3^x-2),2{\log }_9(3^x-8/3)$ are in A.P., then the value of $x$ can be-

• A. $\log \left(\frac{4}{3}\right)$
• B. $\frac{\log 4}{\log 3}$
• C. $1$
• D. ${\log }_{3}4$

Answer 1

The correct option is D ${\log }_{3}4$

We know if $x,y,z$ are in $A.P.$ then $2y=x+z$

So, we given $1,{\log }_3(3^x-2),2{\log }_9(3^x-8_3)$ are in $A.P$

Then $2{\log }_3(3^x-2)=2{\log }_9(3^x-8_3)+1$

Note that ${\log }_9(3^x-8_3)=\frac{1}{2}{\log }_3(3^x-8_3)$

So $2{\log }_3(3^x-2)=\frac{2}{2}{\log }_3(3^x-8_3)+{\log }_3^3$

$2{\log }_3(3^x-2)={\log }_{3}3(3^x-8_3)$

Let $3^x=y2{\log }_3(y-2)={\log }_{3}y-8_3$

$\Rightarrow (y-2{)}^2={\log }_{3}3(y-8_3)$

$\Rightarrow y^2-7y+12=0$

$\Rightarrow y=3$ or $y=4$

$\Rightarrow 3^x=3$ or $3^x=4$

$\Rightarrow x={\log }_3^3$ or $x={\log }_3^4$

$\Rightarrow x=y$ or $x{\log }_3^4$