Find the value of x if 1,$\log_{9}(3^{1-x}+2) $, $ \log_{3}(4.3^x-1) $ are in an AP.

Question

We have an AP: 1, $ \log_{9}(3^{1-x}+2) $, $ \log_{3}(4.3^x-1) $ We have to find value of x. $$ d = a_{2} - a_{1} $$ $$ d = log^{(3^{1-x}+2)}_9 - 1 $$ $$ d = log^{(3^{1-x}+2)^{\frac{1}{2}}}_3 - log^3_3 $$ $$ d = log^{\frac{(3^{1-x}+2)^{\frac{1}{2}}}{3}}_3 $$

Also $$ d = a_{3} - a_{2} $$ $$ d = log^{(4.3^x-1)}_3 - log^{(3^{1-x}+2)^{\frac{1}{2}}}_3 $$ $$ d = log^{\frac{4.3^x-1}{(3^{1-x}+2)^{\frac{1}{2}}}}_3 $$

now

$$ a_{2} - a_{1} = a_{3} - a_{2} $$ $$ log^{\frac{(3^{1-x}+2)^{\frac{1}{2}}}{3}}_3 = log^{\frac{4.3^x-1}{(3^{1-x}+2)^{\frac{1}{2}}}}_3 $$ $$ log^{\frac{(3^{1-x}+2)^{\frac{1}{2}}}{3}}_3 - log^{\frac{4.3^x-1}{(3^{1-x}+2)^{\frac{1}{2}}}}_3 = 0 $$ $$ log^{\frac{(3^{1-x}+2)^{\frac{1}{2}}}{3} . \frac{(3^{1-x}+2)^{\frac{1}{2}}}{4.3^x-1}}_3 = 0 $$ $$ log^{\frac{3^{1-x}+2}{3(4.3^x-1)}}_3 = 0 $$ so $$ \frac{3^{1-x}+2}{3.4.3^x-3} = 1 $$ $$ 3^{1-x}+2 = 3.4.3^x-3 $$ $$ 3.4.3^x-3^{1-x} = 5 $$

I don't know what to do next. Any suggestions?

Answer 1

You have $3.4.3^x-3^{1-x} = 5$. Consider substitution $y=3^{x}$, which results in $12y-\frac{3}{y}=5$, hence $12y^2-3=5y$. Can you proceed from here?

Answer 2

Given $1\;\;,\log_{9}\left(3^{1-x}+2\right)\;\;,\log_{3}\left(4\cdot 3^{x}-1\right)$ are in $\bf{A.P}$

Using the fact that if $a,b,c$ are in $\bf{A.P}\;,$ Then $b-a=c-b\Rightarrow 2b = a+c$

So $$\displaystyle 2\cdot \log_{9}\left(3^{1-x}+2\right)=1+\log_{3}\left(4\cdot3^{x}-1\right) = \log_{3}(3)+\log_{3}\left(4\cdot3^{x}-1\right)$$

so we get $$\displaystyle 2\log_{9}\left(3^{1-x}+2\right)=\ln_{3}[3\cdot (4\cdot 3^x-1)]$$

So $$\displaystyle 2\cdot \frac{1}{2}\log_{3}\left(3^{1-x}+2\right)^2=\ln_{3}[3\cdot (4\cdot 3^x-1)]$$

So we get $$\displaystyle \log_{3}\left(3^{1-x}+2\right)=\ln_{3}[3\cdot (4\cdot 3^x-1)]$$

So $$\displaystyle 3^{1-x}+2 = 3\cdot (4\cdot 3^x-1)\Rightarrow \frac{3}{3^x}+2 = 12\cdot 3^x-3$$

Now Put $3^x = t\;,$ Then equation convert into $$\displaystyle \frac{3}{y}+2 = 12y-3$$

So we get $$\displaystyle 12y^2-5y-3 = 0\Rightarrow 12y^2-9y+4y-3=0$$

So $$\displaystyle 3y(4y-3)-1(4y-3) = 0\Rightarrow (3y-1)\cdot (4y-3) = 0$$

So we get $$\displaystyle y = \frac{1}{3}\Rightarrow 3^x = 3^{-1}\Rightarrow \boxed{x=-1}$$

and $$\displaystyle y=\frac{4}{3}\Rightarrow 3^x = \frac{3}{4} \Rightarrow \log_{3}(3^x) = \log_{3}\left(\frac{3}{4}\right)$$

so we get $$\displaystyle \boxed{x= 1-\log_{3}(4)}$$