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

JEE · Math · previous-year question

  1. A.$${\log _3}\,4\,\,\,$$
  2. B.$$1 - \,{\log _3}\,4\,$$correct
  3. C.$$1 - \,{\log _4}\,3$$
  4. D.$${\log _4}\,3$$

Answer

B. $$1 - \,{\log _3}\,4\,$$

Explanation

$$1,\,{\log _9}\left( {{3^{1 - x}} + 2} \right),{\log _3}\left( {{{4.3}^x} - 1} \right)$$ 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 $$x = {\log _3}3 - {\log _3}4$$ $$ \Rightarrow x = 1 - {\log _3}4$$

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