1. If $$\log{3}, log(3^{x} - 2)$$ and $$log (3^{x}+ 4)$$ are in arithmetic progression, then x is equal to





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  • By: anil on 05 May 2019 03.32 am
    If $$log{3}, log(3^{x} - 2)$$ and $$log (3^{x}+ 4)$$ are in arithmetic progression
    Then, $$2*log(3^{x} - 2) = log{3}+log (3^{x}+ 4)$$
    Thus, $$log{(3^{x} - 2)^2} = log{3(3^x+4)}$$
    Thus, $$(3^{x} - 2)^2 = 3(3^x+4)$$
    => $$3^{2x} - 4*3^x +4 = 3*3^x + 12$$
    => $$3^{2x} - 7*3^x - 8 = 0$$
    => $$(3^x+1)*(3^x-8) = 0$$
    But $$3^x+1
    eq 0$$
    Thus, $$3^x = 8$$
    Hence, $$x = log_{3}{8}$$
    Hence, option B is the correct answer.
  • By: anil on 05 May 2019 03.32 am
    If $$log{3}, log(3^{x} - 2)$$ and $$log (3^{x}+ 4)$$ are in arithmetic progression
    Then, $$2*log(3^{x} - 2) = log{3}+log (3^{x}+ 4)$$
    Thus, $$log{(3^{x} - 2)^2} = log{3(3^x+4)}$$
    Thus, $$(3^{x} - 2)^2 = 3(3^x+4)$$
    => $$3^{2x} - 4*3^x +4 = 3*3^x + 12$$
    => $$3^{2x} - 7*3^x - 8 = 0$$
    => $$(3^x+1)*(3^x-8) = 0$$
    But $$3^x+1
    eq 0$$
    Thus, $$3^x = 8$$
    Hence, $$x = log_{3}{8}$$
    Hence, option B is the correct answer.
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