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.
$${\log _3}4$$
B.
$$1 - {\log _3}4$$
C.
$$1 - {\log _4}3$$
D.
$${\log _4}3$$
Answer :
$$1 - {\log _3}4$$
Solution :
$$\eqalign{
& 1,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log_3} \left( {{{4.3}^x} - 1} \right){\text{ are in A}}{\text{.P}}{\text{.}} \cr
& \Rightarrow \,\,{\text{2lo}}{{\text{g}}_9}\left( {{3^{1 - x}} + 2} \right) = 1 + {\log _3}\left( {{{4.3}^x} - 1} \right) \cr
& \Rightarrow \,\,{\log _3}\left( {{3^{1 - x}} + 2} \right) = {\log _3}3 + {\log _3}\left( {{{4.3}^x} - 1} \right) \cr
& \Rightarrow \,\,{\log _3}\left( {{3^{1 - x}} + 2} \right) = {\log _3}\left[ {3\left( {{{4.3}^x} - 1} \right)} \right] \cr
& \Rightarrow \,\,{3^{1 - x}} + 2 = 3\left( {{{4.3}^x} - 1} \right) \cr
& \Rightarrow \,\,{3.3^{ - x}} + 2 = {12.3^x} - 3. \cr
& {\text{Put }}{{\text{3}}^x} = t \cr
& \Rightarrow \,\,\frac{3}{t} + 2 = 12t - 3{\text{ or 12}}{t^2} - 5t - 3 = 0; \cr
& {\text{Hence }}t = - \frac{1}{3},\frac{3}{4} \cr
& \Rightarrow \,\,{3^x} = \frac{3}{4}\left( {{\text{as }}{{\text{3}}^x} \ne - ve} \right) \cr
& \Rightarrow \,\,x = {\log _3}\left( {\frac{3}{4}} \right){\text{ or }}x = {\log _3}3 - {\log _3}4 \cr
& \Rightarrow \,\,x = 1 - {\log _3}4 \cr} $$