Question

If $1,{\log }_9\left(3^{1-x}+2\right),{\log }_3\left(4.3^x-1\right)$ are in A.P then $x$ equals

• A. ${\log }_3\left(4\right)$
• B. $1-{\log }_3\left(4\right)$
• C. $1-{\log }_4\left(3\right)$
• D. ${\log }_4\left(3\right)$

Answer 1

The correct option is B

$1-{\log }_3\left(4\right)$

The explanation of the correct option:

Step1. Define common differences in Arithmetic progression:

Given, $1,{\log }_9\left(3^{1-x}+2\right),{\log }_3\left(4.3^x-1\right)$ are in A.P.

Let $a_1=1$

$a_2={\log }_9\left(3^{1-x}+2\right)$

$a_3={\log }_3\left(4.3^x-1\right)$

$\therefore {\log }_9\left(3^{1-x}+2\right)-1={\log }_3\left(4.3^x-1\right)-{\log }_9\left(3^{1-x}+2\right)$ $\left[\mathbf{\because }{\mathbf{a}}_{\mathbf{2}}\mathbf{-}{\mathbf{a}}_{\mathbf{1}}\mathbf{=}{\mathbf{a}}_{\mathbf{3}}\mathbf{-}{\mathbf{a}}_{\mathbf{2}}\mathbf{=}\mathbf{d}\right]$

Step2. Convert the logarithm function into an exponential function:

${\log }_9\left(3^{1-x}+2\right)-1={\log }_3\left(4.3^x-1\right)-{\log }_9\left(3^{1-x}+2\right)$

$\Rightarrow$ $2{\log }_9\left(3^{1-x}+2\right)=1+{\log }_3\left(4.3^x-1\right)$

$\Rightarrow$ $2{\log }_{3^2}\left(3^{1-x}+2\right)=1+{\log }_3\left(4.3^x-1\right)$

$\Rightarrow$$2\times \frac{1}{2}{\log }_3\left(3^{1-x}+2\right)=1+{\log }_3\left(4.3^x-1\right)$ $\left[\mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{{\mathbf{a}}^{\mathbf{n}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{n}}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{a}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\right]$

$\Rightarrow$ ${\log }_3\left(3^{1-x}+2\right)={\log }_3\left(3\right)+{\log }_3\left(4.3^x-1\right)$ $\left[\mathbf{\because }{\mathbf{log}}_{\mathbf{a}}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}\mathbf{1}\right]$

$\Rightarrow$ ${\log }_3\left(3^{1-x}+2\right)={\log }_3\left[\left(3\right)\times \left(4.3^x-1\right)\right]$ $\mathbf{[}\mathbf{\because }\mathbf{log}\mathbf{(}\mathbf{a}\mathbf{\times }\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{log}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{+}\mathbf{log}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{]}$

$\Rightarrow$ $3^{1-x}+2=\left(3\right)\times \left(4.3^x-1\right)$

$\Rightarrow$ $3^{1-x}+2=4.3^{x+1}-3$

$\Rightarrow$ $\frac{3}{3^x}+2=\left(4\times 3\right)3^x-3$

$\Rightarrow$ $\left(4\times 3\right)3^x-\frac{3}{3^x}=3+2$

$\Rightarrow$ $\left(12\right)3^x-\frac{3}{3^x}=5$

Step3.Calculate the value of $\text{'}x\text{'}$:

Let $3^x=u$

$\left(12\right)3^x-\frac{3}{3^x}-5=0$

$\Rightarrow$ $12u-\frac{3}{u}-5=0$

$\Rightarrow$ $12u^2-5u-3=0$

$\Rightarrow$ $12u^2-9u+4u-3=0$

$\Rightarrow$$3u\left(4u-3\right)+1\left(4u-3\right)=0$

$\Rightarrow$ $\left(4u-3\right)\left(3u+1\right)=0$

$\Rightarrow$ $4u-3=0or3u+1=0$

$\Rightarrow$ $u=\frac{3}{4}or\frac{-1}{3}$

$\therefore 3^x=\frac{3}{4}$

$\Rightarrow$$x={\log }_3\left(\frac{3}{4}\right)$ $\mathbf{\left[}\mathbf{\because }{\mathbf{a}}^{\mathbf{y}}\mathbf{=}\mathbf{x}\mathbf{\Rightarrow }{\mathbf{log}}_{\mathbf{a}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{y}\mathbf{}\mathbf{and}\mathbf{}\mathbf{log}\mathbf{\left(}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{\right)}\mathbf{=}\mathbf{log}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{-}\mathbf{log}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{\right]}$

$\Rightarrow$$x={\log }_3\left(3\right)-{\log }_3\left(4\right)$

$\mathbf{\therefore }\mathit{x}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{3}}\mathbf{\left(}\mathbf{4}\mathbf{\right)}$

Hence, Option(B) is the correct answer.