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

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

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

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

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

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

$$\,\,\,\,\,\,\,\,\,$$ $$ = {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right]$$

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

$$ \Rightarrow {3.3^{ - x}} + 2 = {12.3^x} - 3.$$

Put $${3^x} = t$$

$$ \Rightarrow {3 \over t} + 2 = 12t - 3$$

or $$12{t^2} - 5t - 3 = 0;$$

Hence $$t = - {1 \over 3},{3 \over 4} \Rightarrow {3^x} = {3 \over 4}$$

(as $${3^x}\,\, \ne \,\, - ve$$ )

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

or $$x = {\log _3}3 - {\log _3}4$$

$$ \Rightarrow x = 1 - {\log _3}4$$