(b) The given number are in $A.P.$

$\therefore 2{\log _9}({3^{1 – x}} + 2) = {\log _3}({4.3^x} – 1) + 1$

$⇒$  $2{\log _{{3^2}}}({3^{1 – x}} + 2) = {\log _3}({4.3^x} – 1) + {\log _3}3$

$⇒$  $\frac{2}{2}{\log _3}({3^{1 – x}} + 2) = {\log _3}[3({4.3^x} – 1)]$

$⇒$  ${3^{1 – x}} + 2 = 3\,({4.3^x} – 1)$

$⇒$  $\frac{3}{y} + 2 = 12y – 3,$ where $y = {3^x}$

$⇒$  $12{y^2} – 5y – 3 = 0$

$y = \frac{{ – 1}}{3}$ or $\frac{3}{4}$

$\Rightarrow {3^x} = \frac{{ – 1}}{3}\,\,$or ${3^x} = \frac{3}{4}$

$x = {\log _3}\,(3/4)$

$ \Rightarrow x = 1 – {\log _3}4$.