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If  1, \log_{9}(3^{1-x}+2), \log_{3}(4.3^x-1)    are in   AP   then  x  equal to

A.   . \log_{3}4               B.       1-\log_{3}4                C.  1-\log_{4}3             D.  \log_{4}3

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@Jeetu

\\ \log_{a}\frac{m}{n}=\log _am-\log_an\\\log_{a^b}m=\frac{1}{b}\log_am\\If\:three\:tern\:are\:in\:AP(let\:a,b\:and\:c)\:then\\b=\frac{a+c}{2}\\Now,come\:to\:the\:question\:\\1,\log_9(3^{1-x}+4)\:and\:\log_3(4.3^x-1)\:are\:in\:AP\\So,2\log_9(3^{1-x}+4)=\log_3(4.3^x-1)+1\\2.\frac{1}{2}\log_3(3^{1-x}+4)=\log_3(4.3^x-1)+1\\\log_3(3^{1-x}+4)-\log_3(4.3^x-1)=1\\\log_3\frac{3^{1-x}+4}{4.3^x-1}=1\\ \frac{3^{1-x}+4}{4.3^x-1}=3\\ \because\:\log_33=1\\I\:hope\:further\:you\:can\:solve\:by\:yourself

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himanshu.meshram

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