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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:
Option: 1
$\log _{3}4$

Option: 2
$1-\log _{3}4$

Option: 3
$1-\log _{4}3$

Option: 4
$\log _{4}3$

Answers (1)

best_answer

As we have learnt,

If a, b, c are in AP, then 2b = a + c

Now,

If 1, $\log_{9}\left ( 3^{1-x}+2 \right ), \log_{3}\left ( 4.3^{x}-1 \right )$ are in AP

so,

$2\log_{9}\left ( 3^{1-x}+2 \right )= 1+ \log_{3}\left ( 4.3^{x} - 1\right )$

$= \log_{3}3+\log_{3}\left ( 4.3^{x} -1\right )$

$\log_{3}\left ( 3^{1-x}+ 2 \right )= \log_{3}3\times \left ( 4.3^{x} -1\right )\;\;\; \left [ \because 2\log_{9}P = {\log_{9^{1/2}}P}=\log_{3}P \right ]$

$3^{1-x}+2= 12.3^{x}-3$

$\Rightarrow \frac{3}{3^{x}}+2 = 12.3^{x}-3$

$\Rightarrow \frac{3}{y}= 12y -5$

$\Rightarrow 3 = 12y^{2}- 5y$

$\Rightarrow 12y^{2} - 5y - 3= 0$

$\Rightarrow 12y^{2} - 9y + 4y - 3 = 0$

$\Rightarrow 3y\left [ 4y-3 \right ]+ 1\left [ 4y-0 \right ]=0$

$\therefore y= -\frac{1}{3} and \ y= \frac{3}{4}$

$\therefore 3^{x}= \frac{3}{4}$

$x=\log_{3}{\frac{3}{4}}$

$x= \log _{3}{3}- \log_{3}{4}=1-\log_{3}{4}$

Posted by

Sayak

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