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Sequences and Series Part 1

Section: Mathematics

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Question : 100 of 100

Marks: +1, -0

If

1, \log_9 (3^{1-x}+2) , \log_3 (4.3^x-1)

are in A.P. then

x

$\log_3 4$
$1-\log_3 4$
$1-\log_4 3$
$\log_4 3$

Solution:

1, \log_9 (3^{1-x}+2) , \log_3 (4.3^x-1)

are in

A . P

.

\Rightarrow 2 \log_9 (3^{1-x}+2)

=1+\log_3 (4.3^x-1)

\Rightarrow \log_3 (3^{1-x}+2)

=\log_3 3+\log_3 (4.3^x-1)

\Rightarrow \log_3 (3^{1-x}+2)

=\log_3 [3 (4.3^x-1)]

\Rightarrow 3^{1-x}+2=3 (4.3^x-1)

\Rightarrow 3.3^{-x}+2=12.3^x-3

Put

3^x=t

\Rightarrow \;\frac{3}{t}+2=12 t-3

or

12 t^2-5 t-3=0

Hence

t=-\;\frac{1}{3}, \;\frac{3}{4} \Rightarrow 3^x=\;\frac{3}{4}

(as

3^x \neq -v e)

\Rightarrow x=\log_3 \left(\;\frac{3}{4}\right)

or

x=\log_3 3-\log_3 4

\Rightarrow x=1-\log_3 4

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